Question

Prove that: - $\left( \sin {{55}^{\circ }}-\sin {{19}^{\circ }} \right)+\left( \sin {{53}^{\circ }}-\sin {{17}^{\circ }} \right)=\cos {{1}^{\circ }}$.

Answer 1

Arrange the terms and group $\sin {{55}^{\circ }}$ and $\sin {{53}^{\circ }}$ together and $\sin {{19}^{\circ }}$ and $\sin {{17}^{\circ }}$ together. Now, apply the formula $\sin A+\sin B=2\sin \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)$ to simplify the terms. Take common terms together and use the formula, $\sin A-\sin B=2\cos \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{A-B}{2} \right)$ for further simplification. Finally, substitute the value, $\sin {{18}^{\circ }}=\dfrac{\sqrt{5}-1}{4}$ to get the answer.

Complete step-by-step solution
Here, we have to prove the expression: -
$\Rightarrow \left( \sin {{55}^{\circ }}-\sin {{19}^{\circ }} \right)+\left( \sin {{53}^{\circ }}-\sin {{17}^{\circ }} \right)=\cos {{1}^{\circ }}$
Let us consider L.H.S., so we have,
$\Rightarrow$ L.H.S. = $\left( \sin {{55}^{\circ }}-\sin {{19}^{\circ }} \right)+\left( \sin {{53}^{\circ }}-\sin {{17}^{\circ }} \right)$
Rearranging the sine terms, we get,
$\Rightarrow$ L.H.S. = $\left( \sin {{55}^{\circ }}+\sin {{53}^{\circ }} \right)-\left( \sin {{19}^{\circ }}+\sin {{17}^{\circ }} \right)$
Applying the identity: - $\sin A+\sin B=2\sin \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)$, we get,
$\Rightarrow$ L.H.S. = $2\sin \left( \dfrac{{{55}^{\circ }}+{{53}^{\circ }}}{2} \right)\cos \left( \dfrac{{{55}^{\circ }}-{{53}^{\circ }}}{2} \right)-2\sin \left( \dfrac{{{19}^{\circ }}+{{17}^{\circ }}}{2} \right)\cos \left( \dfrac{{{19}^{\circ }}-{{17}^{\circ }}}{2} \right)$
$\Rightarrow$ L.H.S. = $2\sin {{54}^{\circ }}\cos {{1}^{\circ }}-2\sin {{18}^{\circ }}\cos {{1}^{\circ }}$
Now, taking common terms together, we get,
$\Rightarrow$ L.H.S. = $2\cos {{1}^{\circ }}\left( \sin {{54}^{\circ }}-\sin {{18}^{\circ }} \right)$
Applying the identity: - $\sin A-\sin B=2\cos \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{A-B}{2} \right)$, we get,
$\Rightarrow$ L.H.S. = $2\cos {{1}^{\circ }}\times 2\cos \left( \dfrac{{{54}^{\circ }}+{{18}^{\circ }}}{2} \right)\sin \left( \dfrac{{{54}^{\circ }}-{{18}^{\circ }}}{2} \right)$
$\Rightarrow$ L.H.S. = $2\cos {{1}^{\circ }}\times 2\cos {{36}^{\circ }}\sin {{18}^{\circ }}$
$\Rightarrow$ L.H.S. = $4\cos {{1}^{\circ }}\times \left( \cos \left( 2\times {{18}^{\circ }} \right)\sin {{18}^{\circ }} \right)$
Using the half angle property of cosine function given by: - $\cos 2\theta =1-2{{\sin }^{2}}\theta$, we get,
$\Rightarrow$ L.H.S. = $4\cos {{1}^{\circ }}\times \left[ \left( 1-2{{\sin }^{2}}{{18}^{\circ }} \right)\times \sin {{18}^{\circ }} \right]$
Substituting the value of $\sin {{18}^{\circ }}=\dfrac{\sqrt{5}-1}{4}$, we get,
$\Rightarrow$ L.H.S. = $4\cos {{1}^{\circ }}\times \left[ \left( 1-2\times {{\left( \dfrac{\sqrt{5}-1}{4} \right)}^{2}} \right)\times \left( \dfrac{\sqrt{5}-1}{4} \right) \right]$
Using the algebraic identity given as: - ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$, we get,
$\Rightarrow$ L.H.S. = 4\cos {{1}^{\circ }}\times \left[ \left\\{ 1-2\times \left( \dfrac{5+1-2\sqrt{5}}{16} \right) \right\\}\times \left( \dfrac{\sqrt{5}-1}{4} \right) \right]
$\Rightarrow$ L.H.S. = 4\cos {{1}^{\circ }}\times \left[ \left\\{ 1-\left( \dfrac{6-2\sqrt{5}}{8} \right) \right\\}\times \left( \dfrac{\sqrt{5}-1}{4} \right) \right]
Taking L.C.M inside the curly bracket, we get,
$\Rightarrow$ L.H.S. = $4\cos {{1}^{\circ }}\times \left[ \left( \dfrac{8-6+2\sqrt{5}}{8} \right)\times \left( \dfrac{\sqrt{5}-1}{4} \right) \right]$
$\Rightarrow$ L.H.S. = $\cos {{1}^{\circ }}\times \left[ \left( \dfrac{2+2\sqrt{5}}{8} \right)\times \left( \sqrt{5}-1 \right) \right]$
$\Rightarrow$ L.H.S. = $\cos {{1}^{\circ }}\times \left[ \left( \dfrac{\sqrt{5}+1}{4} \right)\times \left( \sqrt{5}-1 \right) \right]$
$\Rightarrow$ L.H.S. = $\cos {{1}^{\circ }}\times \dfrac{1}{4}\times \left[ \left( \sqrt{5}+1 \right)\times \left( \sqrt{5}-1 \right) \right]$
Using the identity: - $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$, we get,
$\Rightarrow$ L.H.S. = $\cos {{1}^{\circ }}\times \dfrac{1}{4}\times \left[ {{\left( \sqrt{5} \right)}^{2}}-{{1}^{2}} \right]$
$\Rightarrow$ L.H.S. = $\cos {{1}^{\circ }}\times \dfrac{1}{4}\times \left[ 5-1 \right]$
$\Rightarrow$ L.H.S. = $\cos {{1}^{\circ }}\times \dfrac{1}{4}\times 4$
Canceling the common factors, we get,
$\Rightarrow$ L.H.S. = $\cos {{1}^{\circ }}$ = R.H.S
Hence proved

Note: One may note that we have grouped $\sin {{55}^{\circ }},\sin {{53}^{\circ }}$ and $\sin {{19}^{\circ }},\sin {{17}^{\circ }}$ together. This is done to make the calculation easy. We can also group the terms as given in the question and proceed directly, but then we would have to encounter some hard calculation. Also note that the value of $\sin {{18}^{\circ }},\cos {{18}^{\circ }},\sin {{36}^{\circ }},\cos {{36}^{\circ }},\sin {{54}^{\circ }}$ etc must be remembered. At many places we need their values and it is not given in the question just like the above one.