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Question: Prove that \({\sin ^4}\theta - {\cos ^4}\theta + 1 = 2{\sin ^2}\theta \)...

Prove that sin⁡4θ−cos⁡4θ+1=2sin⁡2θ{\sin ^4}\theta - {\cos ^4}\theta + 1 = 2{\sin ^2}\theta

Explanation

Solution

To solve this question firstly put the value of sin⁡2θ{\sin ^2}\theta in terms of cos⁡2θ{\cos ^2}\theta and after that solve the equation and try to make it equal to R.H.S. The value of sin⁡2θ{\sin ^2}\theta can be obtained from the following trigonometric identity i.e.
⇒sin⁡2θ+cos⁡2θ=1\Rightarrow {\sin ^2}\theta + {\cos ^2}\theta = 1

Complete step-by-step answer:
We have to prove
⇒sin⁡4θ−cos⁡4θ+1=2sin⁡2θ\Rightarrow {\sin ^4}\theta - {\cos ^4}\theta + 1 = 2{\sin ^2}\theta
To prove this equation, we will solve L.H.S and try to make it equal by using the appropriate trigonometric identity.
Solve the L.H.S of the equation i.e.
⇒sin⁡4θ−cos⁡4θ+1=(sin⁡2θ)2−cos⁡4θ+1\Rightarrow {\sin ^4}\theta - {\cos ^4}\theta + 1 = {({\sin ^2}\theta )^2} - {\cos ^4}\theta + 1 …..(1)
Here we will use a trigonometric identity:
⇒sin⁡2θ+cos⁡2θ=1\Rightarrow {\sin ^2}\theta + {\cos ^2}\theta = 1
⇒sin⁡2θ=1−cos⁡2θ\Rightarrow {\sin ^2}\theta = 1 - {\cos ^2}\theta …….(2)
Put this value in equation 1 we get,
⇒sin⁡4θ−cos⁡4θ+1=(1−cos⁡2θ)2−cos⁡4θ+1\Rightarrow {\sin ^4}\theta - {\cos ^4}\theta + 1 = {(1 - {\cos ^2}\theta )^2} - {\cos ^4}\theta + 1
Apply the formula i.e. (a−b)2=a2+b2−2ab{(a - b)^2} = {a^2} + {b^2} - 2ab on (1−cos⁡2θ)2{(1 - {\cos ^2}\theta )^2}we get,
⇒sin⁡4θ−cos⁡4θ+1=1+cos⁡4θ−2cos⁡2θ−cos⁡4θ+1 ⇒L.H.S=1−2cos⁡2θ+1 ⇒L.H.S=2−2cos⁡2θ  \Rightarrow {\sin ^4}\theta - {\cos ^4}\theta + 1 = 1 + {\cos ^4}\theta - 2{\cos ^2}\theta - {\cos ^4}\theta + 1 \\\ \Rightarrow L.H.S = 1 - 2{\cos ^2}\theta + 1 \\\ \Rightarrow L.H.S = 2 - 2{\cos ^2}\theta \\\
Taking 2 common from both the terms we get,
⇒L.H.S=2(1−cos⁡2θ)\Rightarrow L.H.S = 2(1 - {\cos ^2}\theta )
From 2 we get,
⇒2(1−cos⁡2θ)=2sin⁡2θ\Rightarrow 2(1 - {\cos ^2}\theta ) = 2{\sin ^2}\theta=R. H. S
Here you can see that L.H.S = R.H.S
Hence, proved.

Note: Some students take sin⁡θ\sin \theta as product of sin⁡\sin and θ\theta . But it is wrong, the sin⁡θ\sin \theta means sine of angle θ\theta . These trigonometric functions depend only on the value of the angle θ\theta and not on the position chosen on the terminal side of the angle θ\theta .
The trigonometric ratios are the same for the same angles. If the terminal sides coincide with x-axis then cos⁡ecθ&cot⁡θ\cos ec\theta \& \cot \theta are not defined and if it coincides with y axis, then secθ&tan⁡θsec\theta \& \tan \theta are not defined. The trigonometric ratios can be positive and negative depending upon x and y.
An equation which involves trigonometric functions and is true for all those angles for which functions are defined is called a trigonometric identity. Some basic trigonometric identities are as follows: -
sin⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1
1+tan⁡2θ=sec⁡2θ1 + {\tan ^2}\theta = {\sec ^2}\theta
1+cot⁡2θ=cos⁡ec2θ1 + {\cot ^2}\theta = \cos e{c^2}\theta