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Question

Hi, can someone help me in the part highlighted in red

By writing cos 4x = cos (3x+x) and cos 2x = cos (3x-x), show that cos 2x + cos 4x = 2cos3xcosx and cos 2x – cos 4x = 2sin3xsinx.

Given further that cos3xcosx = 0.683 and sin3xsinx = 0.183, find the value of x where 0° < x < 180°.

Thank you very much.

Answer

cos3xcosx = 0.683

2cos3xcosx = 1.366

sin3xsinx = 0.183

2sin3xsinx = 0.366

Applying cos 2x + cos 4x = 2cos3xcosx and cos 2x – cos 4x = 2sin3xsinx:

cos 2x + cos 4x = 1.366   — (1)

cos 2x – cos 4x = 0.366   — (2)

(1)-(2):

2cos4x = 1

cos4x = 1/2

Ref. angle = 60°

Also, 0°<x<180°,

therefore 0°<4x<720°.

4x = 60°, 300°, 420°, 660°

x = 15°, 75°, 105°, 165°

x = 15°, 165° (x = 75°, 105° are inadmissible)

1 Reply 1 Like ✔Accepted Answer

Hi. So sorry again to have deleted your solution by mistake. Thanks for understanding. 🙂

1 Reply 1 Like

Hi SAHMom,

No worries…thanks for helping out.

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