Question

Please help..:)

Answer 1

) Show that the equation
sin(x−60◦) −cos(30◦−x) = 1
can be written in the form cosx = k, where k is a constant.

Answer 2

@Callisto @ganeshie8 @robtobey

Answer 3

* All angles are in degrees.
Left side
= sin(x−60) −cos(30−x)
= cos[90-(x-60)] - cos (30-x)
= cos (150 - x) - cos (30-x)

By cos a - cos b = -2 sin [(a+b)/2)] sin [ (a-b) /2 ]
You can reduce the left in -2cosx + (a constant). Will you try?

Answer 4

cos a - cos b = -2 sin [(a+b)/2)] sin [ (a-b) /2 ]? hw is this possible

Answer 5

can u expand sin(x-60) = sinxcos60- cosxsin60 and then do it

Answer 6

That would be very time-consuming..

Answer 7

Proof of cos a - cos b = -2 sin [(a+b)/2)] sin [ (a-b) /2 ]
*********************************************************
cos(A+B) = cosAcosB - sinAsinB ---(1)
cos(A-B) = cosAcosB + sinAsinB ---(2)

(1)-(2)
cos(A+B) - cos(A-B) = -2sinAsinB

Let a = A+B and b = A-B
A = (a+b)/2) and B = (a-b)/2
Sub. a = A+B and b = A-B into cos(A+B) - cos(A-B) = -2sinAsinB,
we get
cos a - cos b = -2sin [(a+b)/2] sin [(a-b)/2],

Answer 8

thnks

Answer 9

thanks alot again

Answer 10

cn u help me with integration problem