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Question: If \[\sin \theta =\sin {{15}^{\circ }}+\sin {{45}^{\circ }}\] where \[{{0}^{\circ }} < \theta < {{90...

If sin⁡θ=sin⁡15∘+sin⁡45∘\sin \theta =\sin {{15}^{\circ }}+\sin {{45}^{\circ }} where 0∘<θ<90∘{{0}^{\circ }} < \theta < {{90}^{\circ }}, then θ\theta is equal to

  1. 45∘{{45}^{\circ }}
  2. 54∘{{54}^{\circ }}
  3. 60∘{{60}^{\circ }}
  4. 72∘{{72}^{\circ }}
  5. 75∘{{75}^{\circ }}
Explanation

Solution

In this type of question we have to use the concept of trigonometry. Here, we have to use different formulas of trigonometry such as cos⁡(−θ)=cos⁡θ\cos \left( -\theta \right)=\cos \theta , sin⁡x+sin⁡y=2sin⁡(x+y)2cos⁡(x−y)2\sin x+\sin y=2\sin \dfrac{\left( x+y \right)}{2}\cos \dfrac{\left( x-y \right)}{2}, sin⁡θ=cos⁡(π2−θ)\sin \theta =\cos \left( \dfrac{\pi }{2}-\theta \right) etc. Also we have to use the value of sin⁡30∘\sin {{30}^{\circ }} which is equal to 12\dfrac{1}{2}.

Complete step-by-step solution:
Now, we have to find the value of θ\theta if sin⁡θ=sin⁡15∘+sin⁡45∘\sin \theta =\sin {{15}^{\circ }}+\sin {{45}^{\circ }} and 0∘<θ<90∘{{0}^{\circ }} < \theta < {{90}^{\circ }}
Let us consider the equation
⇒sin⁡θ=sin⁡15∘+sin⁡45∘\Rightarrow \sin \theta =\sin {{15}^{\circ }}+\sin {{45}^{\circ }}
As we know that, sin⁡x+sin⁡y=2sin⁡(x+y)2cos⁡(x−y)2\sin x+\sin y=2\sin \dfrac{\left( x+y \right)}{2}\cos \dfrac{\left( x-y \right)}{2} we can simplify the right side of the equation as

& \Rightarrow \sin \theta =2\sin \dfrac{\left( {{15}^{\circ }}+{{45}^{\circ }} \right)}{2}\cos \dfrac{\left( {{15}^{\circ }}-{{45}^{\circ }} \right)}{2} \\\ & \Rightarrow \sin \theta =2\sin \dfrac{{{60}^{\circ }}}{2}\cos \dfrac{\left( -{{30}^{\circ }} \right)}{2} \\\ & \Rightarrow \sin \theta =2\sin {{30}^{\circ }}\cos \left( -{{15}^{\circ }} \right) \\\ \end{aligned}$$ Now as we know that, $$\sin {{30}^{\circ }}=\dfrac{1}{2}$$ and $$\cos \left( -\theta \right)=\cos \theta $$, we can rewrite the above equation as $$\Rightarrow \sin \theta =2\left( \dfrac{1}{2} \right)\cos \left( {{15}^{\circ }} \right)$$ $$\Rightarrow \sin \theta =\cos \left( {{15}^{\circ }} \right)$$ Now, we can rewrite $${{15}^{\circ }}$$ as $${{15}^{\circ }}=\left( {{90}^{\circ }}-{{75}^{\circ }} \right)$$ that is $${{15}^{\circ }}=\left( \dfrac{\pi }{2}-{{75}^{\circ }} \right)$$ and hence we have $$\Rightarrow \cos {{15}^{\circ }}=\cos \left( \dfrac{\pi }{2}-{{75}^{\circ }} \right)$$ Thus we can write the above equation as $$\Rightarrow \sin \theta =\cos \left( \dfrac{\pi }{2}-{{75}^{\circ }} \right)\cdots \cdots \cdots \left( i \right)$$ But we have the relation $$\sin \theta =\cos \left( \dfrac{\pi }{2}-\theta \right)$$ and hence we get $$\Rightarrow \cos \left( \dfrac{\pi }{2}-{{75}^{\circ }} \right)=\sin {{75}^{\circ }}$$ Hence, we can rewrite equation $$\left( i \right)$$ as $$\Rightarrow \sin \theta =\sin {{75}^{\circ }}$$ $$\Rightarrow \theta ={{75}^{\circ }}$$ Hence, the value of $$\theta $$ is $${{75}^{\circ }}$$ if $$\sin \theta =\sin {{15}^{\circ }}+\sin {{45}^{\circ }}$$ **Thus, option (5) is the correct option.** **Note:** In this type of question students have to remember the values of $$\sin \theta $$ for some particular values of $$\theta $$ such as $${{30}^{\circ }}$$, $${{45}^{\circ }}$$, $${{60}^{\circ }}$$ and $${{90}^{\circ }}$$. Students have to remember formulas of trigonometry for solving such types of programs. If the students are unable to remember all the formulas then at least they have to remember how they can derive one formula from another one so that they can easily solve such types of problems. Also students have to note that the value of $$\theta $$ must be less than $${{90}^{\circ }}$$ as they are provided that $${{0}^{\circ }} < \theta < {{90}^{\circ }}$$.