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Question

If $$\sin\theta +\cos\theta =\sqrt { 2 } \cos\theta$$, then the general value of $$\theta$$ is 


Solution

$$ sin\theta +cos\theta=\sqrt{2}cos\theta$$
sguasing both sides, we get
$$(sin\theta+cos\theta)^{2}=(\sqrt{2}cos\theta)^{2}$$
$$ sin^{2}\theta+cos^{2}\theta+2sin\theta cos\theta = 2cos^{2}\theta.$$
$$ sin^{2}\theta-cos^{2}\theta+2sin\theta cos\theta =0$$
$$-(cos^{2}\theta-sin^{2}\theta)+2sin cos\theta=0$$
$$\Rightarrow -cos2\theta+sin2\theta=0$$                      [$$\therefore $$ $$cos^{2}\theta-sin^{2}\theta=cos2\theta 2sin\theta cos\theta=sin2\theta] $$
$$ \Rightarrow sin 20=cos20 $$
$$\Rightarrow sin20=sin(\frac{\pi }{2}-20)$$
$$\therefore  20=\frac{\pi }{2}-20$$
$$\Rightarrow  20+20=\frac{\pi }{2}$$
$$\Rightarrow 40=\frac{\pi }{2}$$
$$\Rightarrow 0=\frac{\pi }{8}$$

1199649_1347920_ans_7ee18a0368d449b1beb9fadb658870af.JPG

Mathematics

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