Question

can anyone prove this for me plz?​

Answer 1

Answer:

jahakshksyeksskn-mshskeje

Answer 2

4(cos^3 10 + sin^3 20) = 3(cos10 + sin20)

Lets use cubes formula: a^3 + b^3 = (a+b)(a^2 - ab + b^2)

4(cos10 + sin20)(cos^2 10 - cos10sin20 + sin^2 20) = 3(cos10 + sin20)

We can now divide both sides from: (cos10 + sin20)

4(cos^2 10 - cos10sin20 + sin^2 20) = 3

Lets use reduction for: cos^2 10 = sin^2 80

Using multiplication formula for [-cos10sin20]

-cos10sin20 = -1/2 (sin10 + sin30) = -1/2(sin10 + 1/2) = -1/2 sin10 - 1/4

It now looks like this:

4(sin^2 80 + sin^2 20 - 1/2 sin10 - 1/4) = 3

sin^2 80 + sin^2 20 - 1/2 sin10 - 1/4 = 3/4

sin^2 80 + sin^2 20 - 1/2 sin10 = 1

Lets remember angle reduction formula:

sin^2 a = (1-cos2a)/2

For: sin^2 80 = (1-cos160)/2

For: sin^2 20 = (1-cos40)/2

(1-cos160)/2 + (1-cos40)/2 - 1/2 sin10 = 1

Multiply both sides from 2:

1 - cos 160 + 1 - cos40 - sin10 = 2

-cos160 - cos40 - sin10 = 0 (multiply by -1)

cos160 + cos40 + sin10 = 0

Cos sum formula: cosa + cosb = 2cos(a+b)/2 * cos(a-b)/2

2cos100cos60 + sin10 = 0

cos100 + sin10 = 0

cos(90+10) + sin10=0

-sin10 + sin10 = 0

0 = 0

since: cos100 is in the 2nd quarter where it's negative.

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