please help me to do this question it is very urgent please please please question 12 &13
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we have
cos 45°
we can write it as
cos(30°+45°)
now it is in the form of cos(A+B)
now we know
cos(A+B)= cosAcosB-sinASinB
= cos30cos45-sin30sin45
= √3/2×1/√2-1/2×1/√2
= √3/2√2-1/2√2
= √3-1/2√2.
hence cos75°= √3-1/2√2
answer 13 th question
we have
cos²60°.tan²30°+sin30°cos30°sin60°tan45°
(1/2)² . (1/√3)² + 1/2 . √3/2 . √3/2 .1
1/12+ 1/8
2+3/24
5/4 ans.
cos 45°
we can write it as
cos(30°+45°)
now it is in the form of cos(A+B)
now we know
cos(A+B)= cosAcosB-sinASinB
= cos30cos45-sin30sin45
= √3/2×1/√2-1/2×1/√2
= √3/2√2-1/2√2
= √3-1/2√2.
hence cos75°= √3-1/2√2
answer 13 th question
we have
cos²60°.tan²30°+sin30°cos30°sin60°tan45°
(1/2)² . (1/√3)² + 1/2 . √3/2 . √3/2 .1
1/12+ 1/8
2+3/24
5/4 ans.
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