### 6 Answers

- PinkgreenLv 71 month ago
sin(x)=7/25, 0<x<pi/2

cos(x-3pi/2)

=

cos[-(3pi/2-x)]

=

cos(3pi/2-x)

{note that cos(-x)=cos(x)}

=

-sin(x)

=

-7/25.

Alternatively,

cos(x-3pi/2)

=

cos(x)cos(3pi/2)+sin(x)sin(3pi/2)

=

0+sin(x)(-1)

=

-sin(x)

=

-7/25.

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- ComoLv 71 month ago
cos ( x - 3π/2 ) = cos x cos 3 π/2 + sin x sin 3 π/2

cos ( x - π/2 ) = - sin x = -7/25

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- PhilipLv 61 month ago
x is in (0,pi/2). sinx = (7/25). Now cos(x -3pi/2) = cos[(x -3pi/2) +2pi] =

cos(x +pi/2) = -sinx = -(7/25).

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- alexLv 71 month ago
cos(x - (3pi/2))=cos(x -(3pi/2)+2pi) = cos(x+(pi/2)) = -sinx = -7/25

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- JohnathanLv 71 month ago
By the Pythagorean Theorem, 7^2 + 24^2 = 25^2, so if sin(x) = 7/25, cos(x) = 24/25. sin(3π/2) and cos(3π/2) are easy if you know the unit circle. Both of the former values will stay positive as x is in quadrant 1. Then utilize this formula: cos(x - y) = cos(x)cos(y) + sin(x)sin(y).

cos(x - 3π/2)

= cos(x)cos(3π/2) + sin(x)sin(3π/2)

= (24/25)(0) + (7/25)(-1)

= 0 - 7/25

= -7/25. Final.

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