Question

Find the value of,
i. ${{\sin }^{2}}82{{\dfrac{1}{2}}^{0}}-{{\sin }^{2}}22{{\dfrac{1}{2}}^{0}}$
ii. ${{\cos }^{2}}112{{\dfrac{1}{2}}^{0}}-{{\sin }^{2}}52{{\dfrac{1}{2}}^{0}}$
iii. ${{\sin }^{2}}52{{\dfrac{1}{2}}^{0}}-{{\sin }^{2}}22{{\dfrac{1}{2}}^{0}}$

Answer 1

In this question, we have been asked to find the value of some trigonometric functions. In order to solve the question, first we need to examine the angles that are given in the question. Then we put the values of that particular angle using trigonometric ratios table and if no value of that particular angle is given in the table then we split the angles into two parts. Then applying trigonometric identities, we will find the value of that given function in the question.

Formula used:
${{\sin }^{2}}a-{{\sin }^{2}}b=\sin \left( a+b \right)\times \sin \left( a-b \right)$
${{\cos }^{2}}a-{{\sin }^{2}}b=\cos \left( a+b \right)\times \cos \left( a-b \right)$

Complete step by step solution:
i. We have given,
${{\sin }^{2}}82{{\dfrac{1}{2}}^{0}}-{{\sin }^{2}}22{{\dfrac{1}{2}}^{0}}$
$\Rightarrow {{\sin }^{2}}82{{\dfrac{1}{2}}^{0}}-{{\sin }^{2}}22{{\dfrac{1}{2}}^{0}}={{\sin }^{2}}{{82.5}^{0}}-{{\sin }^{2}}{{22.5}^{0}}$
Using the identity,
${{\sin }^{2}}a-{{\sin }^{2}}b=\sin \left( a+b \right)\times \sin \left( a-b \right)$
Applying in the above given function, we obtain
$\Rightarrow \sin \left( 82.5+22.5 \right)\times \sin \left( 82.5-22.5 \right)$
$\Rightarrow \sin \left( {{105}^{0}} \right)\times \sin \left( {{60}^{0}} \right)$
Using the trigonometric ratios, putting the values, we get
$\Rightarrow \left( \dfrac{\sqrt{3}+1}{2\sqrt{2}} \right)\times \dfrac{\sqrt{3}}{2}$
Simplifying the above, we get
$\Rightarrow \dfrac{3+\sqrt{3}}{4\sqrt{2}}$
$\therefore {{\sin }^{2}}82{{\dfrac{1}{2}}^{0}}-{{\sin }^{2}}22{{\dfrac{1}{2}}^{0}}=\dfrac{3+\sqrt{3}}{4\sqrt{2}}$

ii. We have given,
${{\cos }^{2}}112{{\dfrac{1}{2}}^{0}}-{{\sin }^{2}}52{{\dfrac{1}{2}}^{0}}$
$\Rightarrow {{\cos }^{2}}112{{\dfrac{1}{2}}^{0}}-{{\sin }^{2}}52{{\dfrac{1}{2}}^{0}}={{\cos }^{2}}{{112.5}^{0}}-{{\sin }^{2}}{{52.5}^{0}}$
Using the identity,
${{\cos }^{2}}a-{{\sin }^{2}}b=\cos \left( a+b \right)\times \cos \left( a-b \right)$
Applying in the above given function, we obtain
$\Rightarrow \cos \left( 112.5+52.5 \right)\times \cos \left( 112.5-52.5 \right)$
$\Rightarrow \cos \left( {{165}^{0}} \right)\times \cos \left( {{60}^{0}} \right)$
Using trigonometric identities, we obtain
$\Rightarrow \cos \left( {{165}^{0}} \right)\times \cos \left( {{60}^{0}} \right)=\cos \left( {{180}^{0}}-{{15}^{0}} \right)\times \cos \left( {{60}^{0}} \right)$
$\Rightarrow \cos \left( {{180}^{0}}-{{15}^{0}} \right)\times \cos \left( {{60}^{0}} \right)=-\cos \left( {{15}^{0}} \right)\times \cos \left( {{60}^{0}} \right)$
Using the trigonometric ratios, putting the values, we get
$\Rightarrow \left( -\dfrac{1+\sqrt{3}}{2\sqrt{2}} \right)\times \dfrac{1}{2}$
($\because \cos {{15}^{0}}=\sin {{75}^{0}}=\dfrac{\sqrt{3}+1}{2\sqrt{2}}$)
Simplifying the above, we get
$\Rightarrow -\dfrac{1+\sqrt{3}}{4\sqrt{2}}$
$\therefore {{\cos }^{2}}112{{\dfrac{1}{2}}^{0}}-{{\sin }^{2}}52{{\dfrac{1}{2}}^{0}}=-\dfrac{1+\sqrt{3}}{4\sqrt{2}}$

iii. We have given,
${{\sin }^{2}}52{{\dfrac{1}{2}}^{0}}-{{\sin }^{2}}22{{\dfrac{1}{2}}^{0}}$

$\Rightarrow {{\sin }^{2}}52{{\dfrac{1}{2}}^{0}}-{{\sin }^{2}}22{{\dfrac{1}{2}}^{0}}={{\sin }^{2}}{{52.5}^{0}}-{{\sin }^{2}}{{22.5}^{0}}$
Using the identity,
${{\sin }^{2}}a-{{\sin }^{2}}b=\sin \left( a+b \right)\times \sin \left( a-b \right)$
Applying in the above given function, we obtain
$\Rightarrow \sin \left( 52.5+22.5 \right)\times \sin \left( 52.5-22.5 \right)$
$\Rightarrow \sin \left( {{75}^{0}} \right)\times \sin \left( {{30}^{0}} \right)$
Applying the trigonometric identities, we get
$\Rightarrow \sin \left( {{45}^{0}}+{{30}^{0}} \right)\times \sin \left( {{30}^{0}} \right)$
Using the identity,
$\sin \left( a+b \right)=\sin a\times \cos b+\cos a\times \sin b$
$\Rightarrow \left( \sin {{45}^{0}}\times \cos {{30}^{0}}+\cos {{45}^{0}}\times \sin {{30}^{0}} \right)\times \sin {{30}^{0}}$
Using the trigonometric ratios, putting the values, we get
$\Rightarrow \left( \dfrac{\sqrt{3}}{2\sqrt{2}}+\dfrac{1}{2\sqrt{2}} \right)\times \dfrac{1}{2}$
Simplifying the above, we get
$\Rightarrow \left( \dfrac{\sqrt{3}+1}{2\sqrt{2}} \right)\times \dfrac{1}{2}$
Simplifying the above, we get
$\Rightarrow \dfrac{\sqrt{3}+1}{4\sqrt{2}}$
$\therefore {{\sin }^{2}}52{{\dfrac{1}{2}}^{0}}-{{\sin }^{2}}22{{\dfrac{1}{2}}^{0}}=\dfrac{\sqrt{3}+1}{4\sqrt{2}}$

Note: We should always remember the basic identities of trigonometry, so that we would solve the question easily. When we have to find the value of any particular angles and the values of that angle is not known then we split the angle into two parts in such a way that the separate values of sine angle are known to us i.e. given in the trigonometric ratios table.