Trigonometric Identities Question 8
Question: If $ \sin (\theta +\alpha )=a $ and $ \sin (\theta +\beta )=b, $ then $ \cos 2,(\alpha -\beta )-4ab,\cos (\alpha -\beta ) $ is equal to
Options:
A) $ 1-a^{2}-b^{2} $
B) $ 1-2a^{2}-2b^{2} $
C) $ 2+a^{2}+b^{2} $
D) $ 2-a^{2}-b^{2} $
Show Answer
Answer:
Correct Answer: B
Solution:
Given that $ \sin ,(\theta +\alpha )=a $ ?..(i) and $ \sin ,(\theta +\beta )=b $ ?..(ii) Now, $ \cos ,(\theta +\alpha )=\sqrt{1-a^{2}},\Rightarrow \theta +\alpha ={{\cos }^{-1}}\sqrt{1-a^{2}} $ and $ \alpha ,-\beta =(\theta +\alpha )-(\theta +\beta ) $ $ ={{\cos }^{-1}}\sqrt{1-a^{2}}-{{\cos }^{-1}}\sqrt{1-b^{2}} $
$ \Rightarrow \alpha -\beta ={{\cos }^{-1}}(\sqrt{1-a^{2}},\sqrt{1-b^{2}}+ab) $
$ \Rightarrow \cos ,(\alpha -\beta )=\sqrt{1-a^{2}},\sqrt{1-b^{2}}+ab $ Now, $ \cos 2,(\alpha -\beta )-4ab\cos ,(\alpha -\beta ) $ $ =2{{\cos }^{2}},(\alpha -\beta )-1-4ab\cos ,(\alpha -\beta ) $ $ =2,{{( \sqrt{1-a^{2}}\sqrt{1-b^{2}}+ab )}^{2}} $ $ -4ab,( \sqrt{1-a^{2}}\sqrt{1-b^{2}}+ab )-1 $ $ =2,{(1-a^{2})(1-b^{2})+a^{2}b^{2}+2ab\sqrt{1-a^{2}}\sqrt{1-b^{2}}} $ $ -4ab,(\sqrt{1-a^{2}}\sqrt{1-b^{2}}+ab) $ $ =2,(1-b^{2}-a^{2}+a^{2}b^{2})+2a^{2}b^{2}-4a^{2}b^{2}-1 $ $ =2,(1-a^{2}-b^{2})-1=1-2a^{2}-2b^{2}. $
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