Question

How do we find the value of $ \sin 45+\cos 60 $ ?

Answer 1

Hint: We start solving the problem by equating the given $ \sin 45+\cos 60 $ to a variable. We then make use of the results $ \sin 45=\dfrac{1}{\sqrt{2}} $ and $ \cos 60=\dfrac{1}{2} $ , to proceed through the problem. We then convert the obtained numbers to the same denominator by performing the necessary calculations. Once we get the denominators same in both the numbers, then we make use of result $ \dfrac{a}{c}+\dfrac{b}{c}=\dfrac{a+b}{c} $ to get the required value of $ \sin 45+\cos 60 $ .

Complete step by step answer:
According to the problem, we are asked to find the value of $ \sin 45+\cos 60 $.
Let us assume $ d=\sin 45+\cos 60 $ ---(1).
We know that $ \sin 45=\dfrac{1}{\sqrt{2}} $ and $ \cos 60=\dfrac{1}{2} $ . Let us use this results in equation (1).
 $ \Rightarrow d=\dfrac{1}{\sqrt{2}}+\dfrac{1}{2} $ ---(2).
Let us convert the denominators of both the terms present in equation (2) equal to make the simplification of sum easier.
 $ \Rightarrow d=\left( \dfrac{1}{\sqrt{2}}\times \dfrac{\sqrt{2}}{\sqrt{2}} \right)+\dfrac{1}{2} $ .
 $ \Rightarrow d=\dfrac{\sqrt{2}}{2}+\dfrac{1}{2} $ ---(3).
We know that the sum of two numbers of form $ \dfrac{a}{c} $ and $ \dfrac{b}{c} $ is defined as $ \dfrac{a}{c}+\dfrac{b}{c}=\dfrac{a+b}{c} $ . Let us use this result in equation (3).
 $ \Rightarrow d=\dfrac{\sqrt{2}+1}{2} $ .
So, we have found the value of $ \sin 45+\cos 60 $ as $ \dfrac{\sqrt{2}+1}{2} $ .
 $ \therefore $ The value of $ \sin 45+\cos 60 $ is $ \dfrac{\sqrt{2}+1}{2} $ .

Note:
We can solve this problem as shown below:
We have $ d=\sin 45+\cos 60 $ .
 $ \Rightarrow d=\sin 45+\cos \left( 90-30 \right) $ ---(4).
We know that $ \cos \left( 90-\theta \right)=\sin \theta $ . Let us use this result in equation (4).
 $ \Rightarrow d=\sin 45+\sin 30 $ ---(5).
We know that $ \sin 45=\dfrac{1}{\sqrt{2}} $ and $ \sin 30=\dfrac{1}{2} $ . Let us use this result in equation (5).
 $ \Rightarrow d=\dfrac{1}{\sqrt{2}}+\dfrac{1}{2} $ .
Now, let us make the denominators equal for both the terms present in d to make the simplification easier.
 $ \Rightarrow d=\left( \dfrac{1}{\sqrt{2}}\times \dfrac{\sqrt{2}}{\sqrt{2}} \right)+\dfrac{1}{2} $ .
 $ \Rightarrow d=\dfrac{\sqrt{2}}{2}+\dfrac{1}{2} $ ---(6).
We know that the sum of two numbers of form $ \dfrac{a}{c} $ and $ \dfrac{b}{c} $ is defined as $ \dfrac{a}{c}+\dfrac{b}{c}=\dfrac{a+b}{c} $ . Let us use this result in equation (6).
 $ \Rightarrow d=\dfrac{\sqrt{2}+1}{2} $ .