##### 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