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Evaluate the following: 2cos^260° + 3sin^245° – 3sin^230° + 2cos^290°

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Question

Evaluate the following:

2cos260° + 3sin245° – 3sin230° + 2cos290°

Evaluate
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Solution

On substituting the values of various T-ratios, we get:

`2cos^2 60^circ + 3 sin^2 45^circ - 3 sin^2 30^circ + 2 cos^2 90^circ`

=`2xx(1/2)^2 + 3 xx(1/sqrt(2))^2 -3 xx (1/2)^2 + 2 xx (0)^2`

=`2xx1/4+3xx1/2-3xx1/4+0`

=`(1/2 +3/2-3/4)=((2+6-3)/4)=5/4`

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Chapter 11: T-Ratios of Some Particular Angles - EXERCISE 11 [Page 572]

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R.S. Aggarwal Mathematics [English] Class 10
Chapter 11 T-Ratios of Some Particular Angles
EXERCISE 11 | Q 6. | Page 572
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