Question

What is \[\sin {{90}^{\circ }}+\sin {{135}^{\circ }}\] ?

Answer 1

Hint: We can solve the given problem by finding the value of each of the terms individually first and then perform addition of the values we get from the both terms. From basic trigonometry we know that the value of $\sin {{90}^{\circ }}$ and we obtain the value of $\sin {{135}^{\circ }}$ by rewriting ${{135}^{\circ }}$ as sum of the known angles and convert it accordingly by using the trigonometric ratios of compound angles formula. Upon getting that value we add it to the other term which is the value of $\sin {{90}^{\circ }}$ .

Complete step by step answer:
The expression we are given is
\[\sin {{90}^{\circ }}+\sin {{135}^{\circ }}\]
Now to complete the summation we must first find the values of both the terms $\sin {{90}^{\circ }}$ and $\sin {{135}^{\circ }}$ individually.
For $\sin {{90}^{\circ }}$ we recall the values of basic trigonometric functions and get that
 $\Rightarrow \sin {{90}^{\circ }}=1$
On the other hand, to find the value of $\sin {{135}^{\circ }}$ we rewrite ${{135}^{\circ }}$ as sum of the known angles i.e., ${{135}^{\circ }}=90+45$
From standard trigonometric formulas of compound angles, we know that
$\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B$
Replacing $A$ by ${{90}^{\circ }}$ and $B$ by \[{{45}^{\circ }}\] in the above formula we get
$\Rightarrow \sin {{135}^{\circ }}=\sin {{90}^{\circ }}\cos {{45}^{\circ }}+\cos {{90}^{\circ }}\sin {{45}^{\circ }}$
From values of basic trigonometric functions, we know that
$\sin {{90}^{\circ }}=1$ , $\cos {{90}^{\circ }}=0$ , $\sin {{45}^{\circ }}=\dfrac{1}{\sqrt{2}}$ and $\cos {{45}^{\circ }}=\dfrac{1}{\sqrt{2}}$ Substituting the above values in the main equation we get
$\Rightarrow \sin {{135}^{\circ }}=1\cdot \dfrac{1}{\sqrt{2}}+0\cdot \dfrac{1}{\sqrt{2}}$
Further simplifying the above equation, we get
\[\Rightarrow \sin {{135}^{\circ }}=\dfrac{1}{\sqrt{2}}\]
Now, we add these obtained values of the terms $\sin {{90}^{\circ }}$ and $\sin {{135}^{\circ }}$ as shown below
\[\sin {{90}^{\circ }}+\sin {{135}^{\circ }}\]
\[=1+\dfrac{1}{\sqrt{2}}\]
\[=\dfrac{\sqrt{2}+2}{2}\]

Therefore, the required given is \[\dfrac{\sqrt{2}+2}{2}\] .

Note: We must be very careful about replacing the trigonometric functions with their values, as students often get confused between the values of the basic trigonometric functions. Instead of applying the formulas of trigonometric ratios of compound angles and expanding it we can also solve the problem by using the trigonometric ratios of allied angles and convert the same accordingly.