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Question: Prove the following equation: \(\sqrt{2}\sin \left( \dfrac{\pi }{4}-\theta \right)=\cos \theta -\sin...

Prove the following equation: 2sin⁡(π4−θ)=cos⁡θ−sin⁡θ\sqrt{2}\sin \left( \dfrac{\pi }{4}-\theta \right)=\cos \theta -\sin \theta

Explanation

Solution

We need to prove the LHS and RHS to be equal. To do this, we will first take the LHS and solve it by the formula sin⁡(x−y)=sin⁡xcos⁡y−cos⁡xsin⁡y\sin \left( x-y \right)=\sin x\cos y-\cos x\sin y and also require standard values like sin⁡π4=12\sin \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}} and cos⁡π4=12\cos \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}} . By solving the whole LHS, we will try to bring it in the form of RHS. Thus, the required condition will be proved.

Complete step-by-step solution:
We will take the LHS and solve it and try to bring it in the form of RHS.
Now, we know that for any two angles ‘x’ and ‘y’ sin⁡(x−y)\sin \left( x-y \right) is given by the formula:
sin⁡(x−y)=sin⁡xcos⁡y−cos⁡xsin⁡y\sin \left( x-y \right)=\sin x\cos y-\cos x\sin y
Now, the LHS is given as:
2sin⁡(π4−θ)\sqrt{2}\sin \left( \dfrac{\pi }{4}-\theta \right)
Now, we will solve the sin⁡(π4−θ)\sin \left( \dfrac{\pi }{4}-\theta \right) part of the LHS by the above mentioned formula.
Here, we have x=π4x=\dfrac{\pi }{4} and y=θy=\theta
Thus, the value of sin⁡(π4−θ)\sin \left( \dfrac{\pi }{4}-\theta \right) will be calculated as:
sin⁡(π4−θ)=sin⁡π4cos⁡θ−cos⁡π4sin⁡θ\sin \left( \dfrac{\pi }{4}-\theta \right)=\sin \dfrac{\pi }{4}\cos \theta -\cos \dfrac{\pi }{4}\sin \theta ……(i)
We know that sin⁡π4=12\sin \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}} and cos⁡π4=12\cos \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}
Thus, putting in these values in equation (i) we get:
sin⁡(π4−θ)=sin⁡π4cos⁡θ−cos⁡π4sin⁡θ ⇒sin⁡(π4−θ)=12cos⁡θ−12sin⁡θ \begin{aligned} & \sin \left( \dfrac{\pi }{4}-\theta \right)=\sin \dfrac{\pi }{4}\cos \theta -\cos \dfrac{\pi }{4}\sin \theta \\\ & \Rightarrow \sin \left( \dfrac{\pi }{4}-\theta \right)=\dfrac{1}{\sqrt{2}}\cos \theta -\dfrac{1}{\sqrt{2}}\sin \theta \\\ \end{aligned}
⇒sin⁡(π4−θ)=12(cos⁡θ−sin⁡θ)\Rightarrow \sin \left( \dfrac{\pi }{4}-\theta \right)=\dfrac{1}{\sqrt{2}}\left( \cos \theta -\sin \theta \right) ……(ii)
Now, if we multiply the value we obtained for sin⁡(π4−θ)\sin \left( \dfrac{\pi }{4}-\theta \right) by 2\sqrt{2} , we will get the value of LHS.
Thus, multiplying equation (ii) by 2\sqrt{2} we get our LHS as:
LHS=2sin⁡(π4−θ) ⇒LHS=2(12(cos⁡θ−sin⁡θ)) ⇒LHS=1(cos⁡θ−sin⁡θ) ⇒LHS=cos⁡θ−sin⁡θ \begin{aligned} & LHS=\sqrt{2}\sin \left( \dfrac{\pi }{4}-\theta \right) \\\ & \Rightarrow LHS=\sqrt{2}\left( \dfrac{1}{\sqrt{2}}\left( \cos \theta -\sin \theta \right) \right) \\\ & \Rightarrow LHS=1\left( \cos \theta -\sin \theta \right) \\\ & \Rightarrow LHS=\cos \theta -\sin \theta \\\ \end{aligned}
We have our RHS as:
RHS=cos⁡θ−sin⁡θRHS=\cos \theta -\sin \theta
This is the same as LHS we obtained.
Thus, LHS=RHSLHS=RHS
Hence, proved.

Note: Take care while taking the angles ‘x’ and ‘y’. They should not be combined when the formula is applied because sin⁡(x−y)\sin \left( x-y \right) is not equal to sin⁡(y−x)\sin \left( y-x \right) . The relation between them is given as sin⁡(x−y)=−sin⁡(y−x)\sin \left( x-y \right)=-\sin \left( y-x \right) . Therefore, care should be taken while applying the formula.