Question

Prove that:-



Cos 10° Cos 30° Cos 50° Cos 70°=3/16


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Answer 1

Step-by-step explanation:

Answers

Praneethworldtopper

Praneethworldtopper

(A) By the product-to-sum formula for cosine:

cos(A)cos(B) = (1/2)[cos(A + B) + cos(A - B)].

So, we have:

cos(10°)cos(30°)cos(50°)cos(70°)

= [cos(70°)cos(10°)][cos(50°)cos(30°)]

= (1/2)(1/2)[cos(80°) + cos(60°)][cos(80°) + cos(20°)], from above

= (1/4)[cos^2(80°) + cos(80°)cos(20°) + cos(80°)cos(60°) + cos(60°)cos(20°)].

Using the above formula again:

(1/4)[cos^2(80°) + cos(80°)cos(20°) + cos(80°)cos(60°) + cos(60°)cos(20°)]

= (1/4)(1/2)[cos(160°) + cos(100°) + cos(60°) + cos(140°) + cos(20°) + cos(80°) + cos(40°) + 1]

= (1/8)[cos(160°) + cos(100°) + cos(60°) + cos(140°) + cos(20°) + cos(80°) + cos(40°) + 1]

= (1/8)[cos(160°) + cos(100°) + cos(140°) + cos(20°) + cos(80°) + cos(40°)] + 3/16.

We now want to show that:

cos(160°) + cos(100°) + cos(140°) + cos(20°) + cos(80°) + cos(40°) = 0.

To do this, re-arrange the terms as follows:

[cos(160°) + cos(20°)] + [cos(140°) + cos(40°)] + [cos(100°) + cos(80°)].

Using the sum-to-product formula:

cos(A)cos(B) = 2cos[(A + B)/2]cos[(A - B)/2],

each bracketed term equals zero (as (A + B)/2 = 90° and cos(90°) = 0), so this equals:

0 + 0 + 0 = 0, as required.

Therefore, cos(10°)cos(30°)cos(50°)cos(70°) = (1/8)(0) + 3/16 = 3/16.

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Answer 2

Answer:

L.H.S. = R.H.S.

 

Step-by-step explanation:

Given :

$\large cos \ 10 \ cos \ 30 \ cos \ 50 \ cos \ 70=\dfrac{3}{16}$

We know value of

$\large cos \ 30=\dfrac{\sqrt3}{2} \ and \ cos \ 120=\dfrac{-1}{2}$

Here we will use these two value

$\large \text{ L.H.S.=cos \ 10 \ cos \ 30 \ cos \ 50 \ cos \ 70 put cos 30 value }\\\\\\\large \text{ L.H.S.=\dfrac{\sqrt3}{2} cos \ 10 \ cos \ 50 \ cos \ 70 Multiply and divide by 2 here}\\\\\\\large L.H.S.=\dfrac{\sqrt3}{2} cos \ 10 \ \dfrac{1}{2}(2 \ cos \ 50 \ cos \ 70)\\\\\\\large \text{now using formula (2 \ cos \ A \ cos \ B)=cos(A+B)+cos(A-B) }\\\\\\\large L.H.S.=\dfrac{\sqrt3}{2} cos \ 10 \ [\dfrac{1}{2}(2 \ cos \ (50+70)+cos(50-70)]$

$\large L.H.S.=\dfrac{\sqrt3}{4} cos \ 10 \ [cos \ 120+cos(20)]\\\\\\\large \ here \ we \ used \ cos(- \theta)=cos (\theta)\\\\\\\large L.H.S.=\dfrac{\sqrt3}{4} cos \ 10 \ [cos \ \dfrac{-1}{2} +cos \ 20]\\\\\\\large we \ can \ write \ cos \ 20 \ as \ cos \ 30-cos \ 10\\\\\\\large L.H.S.=\dfrac{\sqrt3}{4}[\dfrac{-1}{2} \ cos \ 10 \ +\dfrac{-1}{2}(cos \ 10 \ cos \ 20)]$

$\large L.H.S.=\dfrac{\sqrt3}{4}[\dfrac{-1}{2} \ cos \ 10 \ +\dfrac{-1}{2} (cos \ 30-cos \ 10)]\\\\\\\large \text{Here \dfrac{-1}{2} \ cos \ 10 will get cancel out }\\\\\\\large L.H.S.=\dfrac{\sqrt3}{4}[cos \ 30\times \dfrac{1}{2}]\\\\\\\large L.H.S.=\dfrac{\sqrt3}{4}(\dfrac{\sqrt3}{4})\\\\\\\large L.H.S.=\dfrac{3}{16}$

L.H.S. = R.H.S.

Hence proved.

Thus we get answer.