Question

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Chapter name : Compound and multiple angles...


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Answer 1

Trigonometric and algebraic formulas :

1. cosC - cosD = - 2 sin$\frac{C+D}{2}$ sin$\frac{C-D}{2}$

2. a³ + b³ = (a + b) (a² + b² - ab)

3. cos2A = 2 cos²A - 1

4. cos(A + B) + cos(A - B) = 2 cosA cosB

5. cos(90° - A) = sinA

Proof of the problem :

Now, L.H.S. = 4 (cos³20° + cos³40°)

= 4 (cos20° + cos40°)
(cos²20° + cos²40° - cos20° cos40°)

= 2 (cos20° + cos40°)
(2 cos²20° + 2 cos²40° - 2 cos20° cos40°)

= 2 (cos20° + cos40°)
{(cos40° + 1) + (cos80° + 1) - (cos60° + cos20°)}

= 2 (cos20° + cos40°)
{(cos40° + cos80° - cos20°) + 2 - cos60°}

= 2 (cos20° + cos40°)
{(cos80° - cos20°) + cos40° + 2 - 1/2}

= 2 (cos20° + cos40°)
{- 2 sin(80° + 20°)/2 sin(80° - 20°)/2 + cos40° + 3/2}

= 2 (cos20° + cos40°)
{- 2 sin50° (1/2) + cos40° + 3/2}

= 2 (cos20° + cos40°)
{- sin50° + cos40° + 3/2}

= 2 (cos20° + cos40°)
{- sin50° + cos(90° - 50°) + 3/2}

= 2 (cos20° + cos40°)
{- sin50° + sin50° + 3/2}

= 2 (cos20° + cos40°) * (3/2)

= 3 (cos20° + cos40°)

= R.H.S.

Hence, proved.

Answer 2

Answer:

Step-by-step explanation:

We know that:-

cos 3x=4 cos^3(x)-3cos(x)

=>

4cos^3(x)=cos 3x+3 cos(x)

Now,angle 60° lies in first quadrant and 120° in second quadrant.So,cos 60° and cos 120° have Same values with opposite signs, hence they get cancelled.

So,we get the Value of RHS.