Question

Find the value of given trigonometric ratios:
(i) \[\sin {{15}^{o}}\]
(ii) \[\sin {{75}^{o}}\]
(iii) \[\sin {{105}^{o}}\]

Answer 1

Hint: To find the value of \[\sin {{15}^{o}},\] we will use the formula \[\cos 2\theta =1-2{{\sin }^{2}}\theta .\] We will put the value of \[\theta \text{ as }{{15}^{o}}\] and then we will solve it. Also, the second and third part of the question will be solved by using the identity sin (A + B) = sin A cos B + cos A sin B.

Complete step-by-step solution -
In this question given, we have to solve three parts. Thus, we will solve each part one by one. Let us solve the first part.
(i) \[\sin {{15}^{o}}:\] To find the value of \[\sin {{15}^{o}},\] we will use that trigonometric formula in which we can write \[\sin {{15}^{o}}\] in terms of \[{{30}^{o}}\left( \text{as }{{30}^{o}}\text{ is double of }{{15}^{o}} \right).\] So here, we are going to use the formula given below,
\[\cos 2\theta =1-2{{\sin }^{2}}\theta \]
Here, in this formula, we will put the value of \[\theta \text{ as }{{15}^{o}}\] and then we will solve it. Thus, after doing this, we will get the following result,
\[\Rightarrow \cos \left( 2\left( 15 \right) \right)=1-{{\sin }^{2}}\left( 15 \right)\]
\[\Rightarrow \operatorname{co}{{30}^{o}}=1-{{\sin }^{2}}\left( 15 \right)\]
\[\Rightarrow \dfrac{\sqrt{3}}{2}=1-2{{\sin }^{2}}{{15}^{o}}\]
\[\Rightarrow 2{{\sin }^{2}}{{15}^{o}}=1-\dfrac{\sqrt{3}}{2}\]
\[\Rightarrow {{\sin }^{2}}{{15}^{o}}=\dfrac{1}{2}-\dfrac{\sqrt{3}}{4}\]
\[\Rightarrow {{\sin }^{2}}{{15}^{o}}=\dfrac{2-\sqrt{3}}{4}\]
\[\Rightarrow {\sin {{15}^{o}}}=\pm \sqrt{\dfrac{2-\sqrt{3}}{4}}\]
\[\Rightarrow {\sin {{15}^{o}}}=\pm \dfrac{\sqrt{2-\sqrt{3}}}{2}\]
Here, we will consider positive signs only because in the range of \[0\le \theta \le {{180}^{o}},\text{ the }\sin \theta \] is non – negative. Therefore, we get,
\[\Rightarrow \sin {{15}^{o}}=\dfrac{\sqrt{2-\sqrt{3}}}{2}\]
Now, we will multiply with \[\sqrt{2}\] to both the numerator and denominator of the right-hand side. After doing this, we will get,
\[\Rightarrow \sin {{15}^{o}}=\dfrac{\sqrt{2}\left( \sqrt{2-\sqrt{3}} \right)}{2\sqrt{2}}\]
\[\Rightarrow \sin {{15}^{o}}=\dfrac{\sqrt{2\left( 2 \right)-\sqrt{3}\left( 2 \right)}}{2\sqrt{2}}\]
In the above equation, we have applied this identity, \[\sqrt{a}\sqrt{b}=\sqrt{ab}.\]
\[\Rightarrow \sin {{15}^{o}}=\dfrac{\sqrt{4-2\sqrt{3}}}{2\sqrt{2}}\]
Now, we will try to eliminate the under root on the numerator by making it a square term.
\[\Rightarrow \sin {{15}^{o}}=\dfrac{\sqrt{3+1-2\sqrt{3}}}{2\sqrt{2}}\]
\[\Rightarrow \sin {{15}^{o}}=\dfrac{\sqrt{{{\left( \sqrt{3} \right)}^{2}}+{{\left( 1 \right)}^{2}}-2\left( 1 \right)\sqrt{3}}}{2\sqrt{2}}\]
We know that, \[{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab.\] We will use this identity in the above equation,
\[\Rightarrow \sin {{15}^{o}}=\dfrac{\sqrt{{{\left( \sqrt{3}-1 \right)}^{2}}}}{2\sqrt{2}}\]
We know that, \[{{\left( {{a}^{b}} \right)}^{c}}={{a}^{b\times c}}.\] After applying this identity to the above equation, we get,
\[\Rightarrow \sin {{15}^{o}}=\dfrac{{{\left( {{\left( \sqrt{3}-1 \right)}^{2}} \right)}^{\dfrac{1}{2}}}}{2\sqrt{2}}\]
\[\Rightarrow \sin {{15}^{o}}=\dfrac{{{\left( \sqrt{3}-1 \right)}^{2\times \dfrac{1}{2}}}}{2\sqrt{2}}\]
\[\Rightarrow \sin {{15}^{o}}=\dfrac{\sqrt{3}-1}{2\sqrt{2}}\]
(ii) \[\sin {{75}^{o}}:\] To solve this part, we are going to write \[\left( {{75}^{o}} \right)\] as the sum of two angles whose value is known to us. Thus, we can write \[{{75}^{o}}\] as the sum of \[{{30}^{o}}\text{ and }{{45}^{o}}.\] Now, we will apply the identity given below,
sin (A + B) = sin A cos B + cos A sin B
Here, we will put, \[A={{30}^{o}}\text{ and }B={{45}^{o}}\]
\[\Rightarrow \sin \left( {{30}^{o}}+{{45}^{o}} \right)=\sin {{30}^{o}}\cos {{45}^{o}}+\cos {{30}^{o}}\sin {{45}^{o}}\]
\[\Rightarrow \sin \left( {{75}^{o}} \right)=\dfrac{1}{2}\times \dfrac{1}{\sqrt{2}}+\dfrac{\sqrt{3}}{2}\times \dfrac{1}{\sqrt{2}}\]
\[\Rightarrow \sin {{75}^{o}}=\dfrac{1}{2\sqrt{2}}+\dfrac{\sqrt{3}}{2\sqrt{2}}\]
\[\Rightarrow \sin {{75}^{o}}=\dfrac{\sqrt{3}+1}{2\sqrt{2}}\]
(iii) \[\sin {{105}^{o}}:\] We will solve this part similar to the second part by putting the value of \[{{105}^{o}}\] as the sum of two angles. Here, we are going to put \[{{105}^{o}}\] as the sum of \[{{60}^{o}}\text{ and }{{45}^{o}}.\] Thus, by applying the same formula, we get,
\[\sin \left( {{60}^{o}}+{{45}^{o}} \right)=\sin {{60}^{o}}\cos {{45}^{o}}+\cos {{60}^{o}}\sin {{45}^{o}}\]
\[\Rightarrow \sin {{105}^{o}}=\dfrac{\sqrt{3}}{2}\times \dfrac{1}{\sqrt{2}}+\dfrac{1}{2}\times \dfrac{1}{\sqrt{2}}\]
\[\Rightarrow \sin {{105}^{o}}=\dfrac{\sqrt{3}}{2\sqrt{2}}+\dfrac{1}{2\sqrt{2}}\]
\[\Rightarrow \sin {{105}^{o}}=\dfrac{\sqrt{3}+1}{2\sqrt{2}}\].

Note: The value of \[\sin {{75}^{o}}\] can also be obtained as shown.
\[\sin {{75}^{o}}=\sin \left( {{90}^{o}}-{{15}^{o}} \right)=\cos {{15}^{o}}\]
\[\cos {{15}^{o}}=\sqrt{1-{{\sin }^{2}}15}=\sqrt{1-\dfrac{{{\left( \sqrt{3}-1 \right)}^{2}}}{{{\left( 2\sqrt{2} \right)}^{2}}}}=\dfrac{\sqrt{3}+1}{2\sqrt{2}}\]
The value of \[\sin {{105}^{o}}\] can also be obtained in the same way.