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If `sin 2theta+sin 2phi=1//2` and `cos2theta+cos 2phi=3//2`, then `cos^(2)(theta-phi)=`A. `3//8`B. `5//8`C. `3//4`D. `5//4` |
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Answer» Correct Answer - B Given, `sin 2theta + sin 2phi = 1//2` …(1) and `cos 2theta + cos 2phi = 3//2` …(2) Square and adding, `therefore (sin^(2)2theta + cos^(2) 2theta)+(sin^(2)2phi + cos^(2)2phi)+2[sin 2theta sin 2phi + cos 2theta cos 2phi]=1//4 + 9//4` `rArr cos 2theta cos 2phi + sin 2theta sin 2phi = 1//4` `rArr cos (2theta - 2phi)=1//4 rArr cos^(2) (theta - phi) = 5//8` |
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