If left[1-cos left(frac{pi}{2}+alpharight)+si | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given the expression: $\left[1-\cos \left(\frac{\pi}{2}+\alpha\right)+\sin \left(\frac{3 \pi}{2}+\alpha\right)\right]^2+\left[1-\sin \left(\frac{3

Vedclass · Accepted Answer

$52$

Vedclass · Answer

(B) Given the expression: $\left[1-\cos \left(\frac{\pi}{2}+\alpha\right)+\sin \left(\frac{3 \pi}{2}+\alpha\right)\right]^2+\left[1-\sin \left(\frac{3 \pi}{2}-\alpha\right)-\cos \left(\frac{3 \pi}{2}+\alpha\right)\right]^2=a+b \sin ^2\left(\frac{\pi}{4}+\alpha\right)$.Using trigonometric identities: $\cos(\frac{\pi}{2}+\alpha) = -\sin \alpha$,$\sin(\frac{3\pi}{2}+\alpha) = -\cos \alpha$,$\sin(\frac{3\pi}{2}-\alpha) = -\cos \alpha$,and $\cos(\frac{3\pi}{2}+\alpha) = \sin \alpha$.The expression becomes: $(1+\sin \alpha-\cos \alpha)^2+(1+\cos \alpha-\sin \alpha)^2 = a+b \sin ^2(\frac{\pi}{4}+\alpha)$.Expanding both squares: $(1+\sin^2 \alpha+\cos^2 \alpha+2\sin \alpha-2\sin \alpha \cos \alpha-2\cos \alpha) + (1+\cos^2 \alpha+\sin^2 \alpha+2\cos \alpha-2\sin \alpha \cos \alpha-2\sin \alpha) = a+b \sin^2(\frac{\pi}{4}+\alpha)$.Simplifying: $4-4\sin \alpha \cos \alpha = a+b(\sin \frac{\pi}{4} \cos \alpha+\cos \frac{\pi}{4} \sin \alpha)^2$.$4-4\sin \alpha \cos \alpha = a+\frac{b}{2}(\cos \alpha+\sin \alpha)^2 = a+\frac{b}{2}(1+2\sin \alpha \cos \alpha) = (a+\frac{b}{2}) + b\sin \alpha \cos \alpha$.Comparing coefficients: $b = -4$ and $a+\frac{b}{2} = 4$ $\Rightarrow a-2 = 4$ $\Rightarrow a = 6$.Thus,$a^2+b^2 = 6^2+(-4)^2 = 36+16 = 52$.

If $\left[1-\cos \left(\frac{\pi}{2}+\alpha\right)+\sin \left(\frac{3 \pi}{2}+\alpha\right)\right]^2+\left[1-\sin \left(\frac{3 \pi}{2}-\alpha\right)-\cos \left(\frac{3 \pi}{2}+\alpha\right)\right]^2=a+b \sin ^2\left(\frac{\pi}{4}+\alpha\right)$,then $a^2+b^2=$

Similar Questions

If $5 \tan \theta = 4$,then $\frac{5 \sin \theta - 3 \cos \theta}{5 \sin \theta + 2 \cos \theta} = $

The value of $(1+\cos \frac{\pi}{6})(1+\cos \frac{\pi}{3})(1+\cos \frac{2\pi}{3})(1+\cos \frac{7\pi}{6})$ is

If $\theta = \frac{17 \pi}{3}$,then $(\tan \theta - \cot \theta) = \dots$

If $\cosh x = \frac{\sqrt{14}}{3}$,$\sinh x = \cos \theta$ and $-\pi < \theta < -\frac{\pi}{2}$,then $\sin \theta =$

Find the degree measure corresponding to the following radian measure (Use $\pi = \frac{22}{7}$): $\frac{7 \pi}{6}$ (in $^{\circ}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module