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Trigonometric Identities – Examples

Based on what we have explained to the article Trigonometric Identities, we are going to solve some exercises below:

Example 1: Simplify

a) $ \displaystyle \tan \theta \cot \theta $

$ \displaystyle \tan \theta \cot \theta =$$ \displaystyle \frac{{\sin \theta }}{{\cos \theta }}\cdot \frac{{\cos \theta }}{{\sin \theta }}=$$ \displaystyle \frac{{\cancel{{\sin \theta }}}}{{\cancel{{\cos \theta }}}}\cdot \frac{{\cancel{{\cos \theta }}}}{{\cancel{{\sin \theta }}}}=1$

b) $ \displaystyle (1-\sin \theta )(1+\sin \theta )$

$ \displaystyle (1-\sin \theta )(1+\sin \theta )=$$ \displaystyle 1+\sin \theta -\sin \theta -{{\sin }^{2}}\theta =$$ \displaystyle 1+\cancel{{\sin \theta }}-\cancel{{\sin \theta }}-{{\sin }^{2}}\theta =$$ \displaystyle 1-{{\sin }^{2}}\theta ={{\cos }^{2}}\theta $

c) $ \displaystyle \sin \theta \cot \theta $

$ \displaystyle \sin \theta \cot \theta =$$ \displaystyle \sin \theta \cdot \frac{{\cos \theta }}{{\sin \theta }}=$$ \displaystyle \cancel{{\sin \theta }}\cdot \frac{{\cos \theta }}{{\cancel{{\sin \theta }}}}=\cos \theta $

d) $ \displaystyle {{\sin }^{3}}x+{{\cos }^{2}}x\sin x$

$ \displaystyle {{\sin }^{3}}x+{{\cos }^{2}}x\sin x=$$ \displaystyle \sin x({{\sin }^{2}}x+{{\cos }^{2}}x)=$$\displaystyle \sin x\cdot 1=\sin x$

e) $ \displaystyle \tan x+\cot x$

$ \displaystyle \tan x+\cot x=$$ \displaystyle \frac{{\sin x}}{{\cos x}}+\frac{{\cos x}}{{\sin x}}=$$ \displaystyle \frac{{{{{\sin }}^{2}}x+{{{\cos }}^{2}}x}}{{\cos x\sin x}}=$$ \displaystyle \frac{1}{{\cos x\sin x}}$

f) $ \displaystyle \frac{1}{{1+\cos x}}+\frac{1}{{1-\cos x}}$

$ \displaystyle \frac{1}{{1+\cos x}}+\frac{1}{{1-\cos x}}=$$ \displaystyle \frac{{1-\cos x+1+\cos x}}{{(1-cox)(1+\cos x)}}=$$ \displaystyle \frac{2}{{1+{{{\cos }}^{2}}x}}=\frac{2}{{{{{\sin }}^{2}}x}}$$ \displaystyle \frac{2}{{1+{{{\cos }}^{2}}x}}=\frac{2}{{{{{\sin }}^{2}}x}}$

g) $ \displaystyle \frac{{{{{(\sin x+\cos x)}}^{2}}}}{{\sin x}}-2\cos x$

$ \displaystyle \frac{{{{{(\sin x+\cos x)}}^{2}}}}{{\sin x}}-2\cos x=$$ \displaystyle \frac{{{{{\sin }}^{2}}x+2\sin x\cos x+{{{\cos }}^{2}}x}}{{\sin x}}-2\cos x=$$ \displaystyle \frac{{1+2\sin x\cos x}}{{\sin x}}-2\cos x=$

$ \displaystyle \frac{{1+2\sin x\cos x-2\sin x\cos x}}{{\sin x}}=$$ \displaystyle \frac{{1+\cancel{{2\sin x\cos x}}-\cancel{{2\sin x\cos x}}}}{{\sin x}}=$$ \displaystyle \frac{1}{{\sin x}}$

h) $ \displaystyle (\sin x+\cos x\cot x)\tan x=$

$ \displaystyle (\sin x+\cos x\cot x)\tan x=$$ \displaystyle (\sin x+\cos x\frac{{\cos x}}{{\sin x}})\frac{{\sin x}}{{\cos x}}=$$ \displaystyle \sin x\cdot \frac{{\sin x}}{{\cos x}}+\cos x\cdot \frac{{\cos x}}{{\sin x}}\cdot \frac{{\sin x}}{{\cos x}}=$

$ \displaystyle \frac{{{{{\sin }}^{2}}x}}{{\cos x}}+\cos x=$$ \displaystyle \frac{{{{{\sin }}^{2}}x+{{{\cos }}^{2}}x}}{{\cos x}}=$$ \displaystyle \frac{1}{{\cos x}}$

i) $ \displaystyle \cos ({{90}^{\circ }}-\theta )\sin \theta +\sin ({{90}^{\circ }}-\theta )\cos \theta $

$ \displaystyle \cos ({{90}^{\circ }}-\theta )\sin \theta +\sin ({{90}^{\circ }}-\theta )\cos \theta =$$ \displaystyle \sin \theta \sin \theta +\cos \theta \cos \theta =$$ \displaystyle {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$

j) $ \displaystyle 2\cos 2x+\sin 2x\tan x$

$ \displaystyle 2\cos 2x+\sin 2x\tan x=$$ \displaystyle 2({{\cos }^{2}}x-{{\sin }^{2}}x)+2\sin x\cos x\cdot \frac{{\sin x}}{{\cos x}}=$$ \displaystyle 2{{\cos }^{2}}x-2{{\sin }^{2}}x+2\sin x\cancel{{\cos x}}\cdot \frac{{\sin x}}{{\cancel{{\cos x}}}}=$

$ \displaystyle 2{{\cos }^{2}}x-2{{\sin }^{2}}x+2{{\sin }^{2}}x=$$ \displaystyle 2{{\cos }^{2}}x\cancel{{-2{{{\sin }}^{2}}x}}+\cancel{{2{{{\sin }}^{2}}x=}}2{{\cos }^{2}}x$

Example 2: Evaluate

a) $ \displaystyle 2\sin {{90}^{{}^\circ }}+3\cos {{0}^{{}^\circ }}+4\tan {{0}^{{}^\circ }}+5\cot {{270}^{{}^\circ }}$

$ \displaystyle 2\sin {{90}^{{}^\circ }}+3\cos {{0}^{{}^\circ }}+4\tan {{0}^{{}^\circ }}+5\cot {{270}^{{}^\circ }}=$$ \displaystyle 2\cdot 1+3\cdot 1+4\cdot 0+5\cdot 0=$$ \displaystyle 2+3+0+0=5$

b) $ \displaystyle 3\sin {{270}^{{}^\circ }}+2\tan {{180}^{{}^\circ }}+2\cos {{180}^{{}^\circ }}+\cot {{90}^{{}^\circ }}$

$ \displaystyle 3\sin {{270}^{{}^\circ }}+2\tan {{180}^{{}^\circ }}+2\cos {{180}^{{}^\circ }}+\cot {{90}^{{}^\circ }}=$$ \displaystyle 3\cdot (-1)+2\cdot 0+2\cdot (-1)+0=$$ \displaystyle -3+0-2=-5$

c) $ \displaystyle \sqrt{2}\cos \frac{\pi }{4}+2\sqrt{3}\sin \frac{\pi }{3}$

$ \displaystyle \sqrt{2}\cos \frac{\pi }{4}+2\sqrt{3}\sin \frac{\pi }{3}=$$ \displaystyle \sqrt{2}\cdot \frac{{\sqrt{2}}}{2}+2\sqrt{3}\cdot \frac{{\sqrt{3}}}{2}=$$ \displaystyle \frac{2}{2}+\frac{{2\cdot 3}}{2}=1+3=4$

d) $ \displaystyle \sin {{80}^{{}^\circ }}\cos {{350}^{{}^\circ }}-\sin {{190}^{{}^\circ }}\cos {{280}^{{}^\circ }}$

$ \displaystyle \sin {{80}^{{}^\circ }}\cos {{350}^{{}^\circ }}-\sin {{190}^{{}^\circ }}\cos {{280}^{{}^\circ }}=$$ \displaystyle \sin ({{90}^{{}^\circ }}-{{10}^{{}^\circ }})\cos ({{360}^{{}^\circ }}-{{10}^{{}^\circ }})-\sin ({{180}^{{}^\circ }}+{{10}^{{}^\circ }})\cos ({{270}^{{}^\circ }}+{{10}^{{}^\circ }})=$$ \displaystyle \cos {{10}^{{}^\circ }}\cos {{10}^{{}^\circ }}+\sin {{10}^{{}^\circ }}\sin {{10}^{{}^\circ }}=$$ \displaystyle {{\cos }^{2}}{{10}^{{}^\circ }}+{{\sin }^{2}}{{10}^{{}^\circ }}=1$

d) $ \displaystyle \sin {{15}^{{}^\circ }}$

$ \displaystyle \sin {{15}^{{}^\circ }}=\cos ({{90}^{{}^\circ }}-{{15}^{{}^\circ }})=$$ \displaystyle \cos {{75}^{{}^\circ }}=\cos ({{45}^{{}^\circ }}+{{30}^{{}^\circ }})=$$ \displaystyle \cos {{45}^{{}^\circ }}\cos {{30}^{{}^\circ }}-\sin {{45}^{{}^\circ }}\sin {{30}^{{}^\circ }}=$

e) $ \displaystyle \sin {{15}^{{}^\circ }}$

$ \displaystyle \sin {{15}^{{}^\circ }}=\cos ({{90}^{{}^\circ }}-{{15}^{{}^\circ }})=$$ \displaystyle \cos {{75}^{{}^\circ }}=\cos ({{45}^{{}^\circ }}+{{30}^{{}^\circ }})=$$ \displaystyle \cos {{45}^{{}^\circ }}\cos {{30}^{{}^\circ }}-\sin {{45}^{{}^\circ }}\sin {{30}^{{}^\circ }}=$

Example 3: Find the value of the angle x.

a) $ \displaystyle \cos (2x-{{10}^{{}^\circ }})=\sin (2x+{{10}^{{}^\circ }})$

$ \displaystyle \cos ({{90}^{\circ }}-x)=\sin x$

$ \displaystyle \cos (2x-{{10}^{{}^\circ }})=\sin (2x+{{10}^{{}^\circ }})$

$ \displaystyle \cos ({{90}^{\circ }}-(2x-{{10}^{{}^\circ }}))=\sin (2x+{{10}^{{}^\circ }})$

$ \displaystyle {{90}^{\circ }}-2x+{{10}^{{}^\circ }}=2x+{{10}^{{}^\circ }}$

$ \displaystyle {{100}^{{}^\circ }}-2x=2x+{{10}^{{}^\circ }}$

$ \displaystyle 2x+2x={{100}^{{}^\circ }}-{{10}^{{}^\circ }}$

$ \displaystyle 4x={{90}^{\circ }}$

$ \displaystyle x={{22,5}^{{}^\circ }}$

b) $ \displaystyle \log (\sin x)=0$

$ \displaystyle \log (\sin x)=0$$ \displaystyle \sin x={{10}^{0}}$$ \displaystyle \sin x=1$.

An angle which has the sinuses 1 is 90°. So the solutions are: $ \displaystyle x=2k\pi +\frac{\pi }{2}$ or $ \displaystyle x=k{{360}^{{}^\circ }}+{{90}^{\circ }}$

c) $ \displaystyle \frac{{2{{{\cos }}^{2}}x+\cos x}}{{\cos x}}=0$

$ \displaystyle \frac{{\cos x(2\cos x+1)}}{{\cos x}}=0$

$ \displaystyle \frac{{\cancel{{\cos x}}(2\cos x+1)}}{{\cancel{{\cos x}}}}=0$

$ \displaystyle 2\cos x+1=0$

$ \displaystyle 2\cos x=-1$

$ \displaystyle \cos x=-\frac{1}{2}$

An angle which has the cosines $ \displaystyle -\frac{1}{2}$ is 120° and -120°. So the solutions are: $ \displaystyle x=2k\pi \pm \frac{{2\pi }}{3}$ or $ \displaystyle x=k{{360}^{{}^\circ }}\pm {{120}^{{}^\circ }}$

Example 4: Prove the identity.

a) $ \displaystyle \frac{{\tan x{{{\sec }}^{2}}x\sin x}}{{\cos e{{c}^{2}}x{{{\cos }}^{2}}x\cot x}}=1$

$ \displaystyle \frac{{\frac{{\sin x}}{{\cos x}}\cdot \frac{1}{{{{{\cos }}^{2}}x}}\cdot \sin x}}{{\frac{1}{{{{{\sin }}^{3}}x}}\cdot {{{\cos }}^{2}}x\cdot \frac{{\cos x}}{{\sin x}}}}=$$ \displaystyle \frac{{\frac{{\sin x}}{{\cos x}}\cdot \frac{{\sin x}}{{\cos x}}\cdot \frac{1}{{\cos x}}}}{{\frac{{{{{\cos }}^{2}}x}}{{{{{\sin }}^{2}}x}}\cdot \frac{{\cos x}}{{\sin x}}\cdot \frac{1}{{\sin x}}}}=$$ \displaystyle \frac{{\tan x\cdot \tan x\cdot \frac{1}{{\cos x}}}}{{{{{\tan }}^{2}}x\cdot \tan x\frac{1}{{\sin x}}}}=$$ \displaystyle \frac{{{{{\tan }}^{2}}x\cdot \sin x}}{{{{{\tan }}^{3}}x\cdot \cos x}}=$

$ \displaystyle \frac{{{{{\tan }}^{2}}x\cdot \tan x}}{{{{{\tan }}^{3}}x}}=$$ \displaystyle \frac{{{{{\tan }}^{3}}x}}{{{{{\tan }}^{3}}x}}=1$

b) $ \displaystyle \frac{{1+\sin 3x}}{{1-\sin 3x}}-\frac{{1-\sin 3x}}{{1+\sin 3x}}=4\tan 3x\sec 3x$

$ \displaystyle \frac{{(1+\sin 3x)(1+\sin 3x)-(1-\sin 3x)(1-\sin 3x)}}{{(1-\sin 3x)1+\sin 3x}}=$

$ \displaystyle \frac{{(1+2\sin 3x+{{{\sin }}^{2}}3x)-(1-2\sin 3x+{{{\sin }}^{2}}3x)}}{{1-{{{\sin }}^{2}}3x}}=$

$ \displaystyle \frac{{1+2\sin 3x+{{{\sin }}^{2}}3x-1+2\sin 3x-{{{\sin }}^{2}}3x}}{{1-{{{\sin }}^{2}}3x}}=$

$ \displaystyle \frac{{1+2\sin 3x+{{{\sin }}^{2}}3x-1+2\sin 3x-{{{\sin }}^{2}}3x}}{{1-{{{\sin }}^{2}}3x}}=$

$ \displaystyle \frac{{\cancel{1}+2\sin 3x\cancel{{+{{{\sin }}^{2}}3x}}\cancel{{-1}}+2\sin 3x\cancel{{-{{{\sin }}^{2}}3x}}}}{{1-{{{\sin }}^{2}}3x}}=$

$ \displaystyle \frac{{4\sin 3x}}{{{{{\cos }}^{2}}3x}}=\frac{{4\sin 3x}}{{\cos 3x\cos 3x}}=$$ \displaystyle \frac{{4\tan 3x}}{{\cos 3x}}=4\tan 3x\sec 3x$

c) $ \displaystyle \cos ec2x+\cot 2x=\cot x$

$ \displaystyle \frac{1}{{\sin 2x}}+\frac{{\cos 2x}}{{\sin 2x}}=\frac{{1+\cos 2x}}{{\sin 2x}}=$$ \displaystyle \frac{{1+{{{\cos }}^{2}}x-{{{\sin }}^{2}}x}}{{\sin 2x}}=\frac{{2{{{\cos }}^{2}}x}}{{\sin 2x}}=$$ \displaystyle \frac{{2{{{\cos }}^{2}}x}}{{2\sin x\cos x}}=\frac{{\cancel{2}\cancel{{\cos x}}\cos x}}{{\cancel{2}\sin x\cancel{{\cos x}}}}=$$ \displaystyle \frac{{\cos x}}{{\sin x}}=\cot x$

d) $ \displaystyle {{\cos }^{4}}x-{{\sin }^{4}}x={{\cos }^{2}}x-{{\sin }^{2}}x$

$ \displaystyle {{\cos }^{4}}x-{{\sin }^{4}}x=({{\cos }^{2}}x-{{\sin }^{2}}x)({{\cos }^{2}}x+{{\sin }^{2}}x)$

$ \displaystyle {{\cos }^{4}}x-{{\sin }^{4}}x=({{\cos }^{2}}x-{{\sin }^{2}}x)\cdot 1$

$ \displaystyle {{\cos }^{4}}x-{{\sin }^{4}}x=({{\cos }^{2}}x-{{\sin }^{2}}x)$

e) $ \displaystyle tgx+\frac{1}{{tgx}}=\frac{1}{{\sin x\cos x}}$

$ \displaystyle tgx+\frac{1}{{tgx}}=\frac{{\sin x}}{{\cos x}}+\frac{1}{{\frac{{\sin x}}{{\cos x}}}}$

$ \displaystyle tgx+\frac{1}{{tgx}}=\frac{{\sin x}}{{\cos x}}+\frac{{\cos x}}{{\sin x}}$

$ \displaystyle tgx+\frac{1}{{tgx}}=\frac{{(\sin x\cdot \sin x)+(\cos x\cdot \cos x)}}{{\cos x\cdot \sin x}}$

$ \displaystyle tgx+\frac{1}{{tgx}}=\frac{{{{{\sin }}^{2}}x+{{{\cos }}^{2}}x}}{{\cos x\sin x}}$

$ \displaystyle tgx+\frac{1}{{tgx}}=\frac{1}{{\sin x\cos x}}$

f) $\displaystyle \frac{{1-t{{g}^{2}}x}}{{1+t{{g}^{2}}x}}=1-2{{\sin }^{2}}x$

$ \displaystyle \frac{{1-t{{g}^{2}}x}}{{1+t{{g}^{2}}x}}=\frac{{1-\frac{{{{{\sin }}^{2}}x}}{{{{{\cos }}^{2}}x}}}}{{1+\frac{{{{{\sin }}^{2}}x}}{{{{{\cos }}^{2}}x}}}}$

$ \displaystyle \frac{{1-t{{g}^{2}}x}}{{1+t{{g}^{2}}x}}=\frac{{{{{\cos }}^{2}}x-{{{\sin }}^{2}}x}}{{{{{\cos }}^{2}}x+{{{\sin }}^{2}}x}}$

$ \displaystyle \frac{{1-t{{g}^{2}}x}}{{1+t{{g}^{2}}x}}=\frac{{{{{\cos }}^{2}}x-{{{\sin }}^{2}}x}}{1}$

$ \displaystyle \frac{{1-t{{g}^{2}}x}}{{1+t{{g}^{2}}x}}={{\cos }^{2}}x-{{\sin }^{2}}x$

$ \displaystyle \frac{{1-t{{g}^{2}}x}}{{1+t{{g}^{2}}x}}=1-{{\sin }^{2}}x-{{\sin }^{2}}x$

$ \displaystyle \frac{{1-t{{g}^{2}}x}}{{1+t{{g}^{2}}x}}=1-2{{\sin }^{2}}x$

g) $ \displaystyle tgx+\cot x=\sec x\cos ecx$

Simplifying both sides at the same time $ \displaystyle \frac{{\sin x}}{{\cos x}}+\frac{{\cos x}}{{\sin x}}=\frac{1}{{\cos x}}\cdot \frac{1}{{\sin x}}$

$ \displaystyle \frac{{{{{\sin }}^{2}}x+{{{\cos }}^{2}}x}}{{\cos x\sin x}}=\frac{1}{{\cos x\sin x}}$

$ \displaystyle \frac{1}{{\cos x\sin x}}=\frac{1}{{\cos x\sin x}}$

h) $ \displaystyle \sec x-tgx=\frac{1}{{\sec x+tgx}}$

Simplifying both sides at the same time $ \displaystyle \frac{1}{{\cos x}}-\frac{{\sin x}}{{\cos x}}=\frac{1}{{\frac{1}{{\cos x}}+\frac{{\sin x}}{{\cos x}}}}$

$ \displaystyle \frac{{1-\sin x}}{{\cos x}}=\frac{1}{{\frac{{1+\sin x}}{{\cos x}}}}$

$ \displaystyle \frac{{1-\sin x}}{{\cos x}}=\frac{{\cos x}}{{1+\sin x}}$

We use the cross multiplication method to prove the identity

$ \displaystyle (1-\sin x)(1+\sin x)=\cos x\cdot \cos x$

$ \displaystyle 1-{{\sin }^{2}}x={{\cos }^{2}}x$

We obtained the basic trigonometric formula $ \displaystyle {{\cos }^{2}}x+{{\sin }^{2}}x=1$

i) $ \displaystyle (1+\sec x)(1-\cos x)=tgx\sin x$

$ \displaystyle (1+\frac{1}{{\cos x}})(1-\cos x)=\frac{{\sin x}}{{\cos x}}\cdot \sin x$

$ \displaystyle (\frac{{1+\cos x}}{{\cos x}})(1-\cos x)=\frac{{{{{\sin }}^{2}}x}}{{\cos x}}$

$ \displaystyle \frac{{1-{{{\cos }}^{2}}x}}{{\cos x}}=\frac{{{{{\sin }}^{2}}x}}{{\cos x}}$

$ \displaystyle \frac{{{{{\sin }}^{2}}x}}{{\cos x}}=\frac{{{{{\sin }}^{2}}x}}{{\cos x}}$

Reminder: To learn Trigonometric identities read Trigonometric Identities

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