Question

14. Prove that : cos10 cos30° cos50° cos70° = 3/16​

Answer 1

Answer:

Step-by-step explanation:

Given:

cos 10°. cos 30°. cos 50°. cos 70° = 3/16

To Prove:

LHS = RHS

Proof:

Consider the LHS of the equation,

cos 10°. cos 30°. cos 50°. cos 70°

We know that,

cos 30° = √3/2

Hence,

⇒ cos 10° × √3/2 × cos 50° × cos 70°

⇒ √3/2 × (cos 50° × cos 10°) × cos 70°

Multiplying and dividing by 2

⇒ √3/2 × 2/2 (cos 50° × cos 10°) × cos 70°

⇒ √3/4 × 2 (cos 50° × cos 10°) × cos 70°

We know that,

2 cosx cosy = cos (x - y) + cos (x + y)

Applying the identity,

⇒ √3/4 × (cos (50° - 10°) + cos (50° + 10°) ) × cos 70°

⇒ √3/4 × ( cos 40° + cos 60°) × cos 70°

We know,

cos 60° = 1/2

⇒ √3/4 × ( cos 40° + 1/2) × cos 70°

⇒ √3/4 × ( cos 40° cos 70° + 1/2 × cos 70°)

Again multiplying and dividing by 2,

⇒ √3/4 × (2/2 × cos 40° cos 70° + 1/2 cos 70°)

Applying the above identity again,

⇒ √3/4 × [1/2 ( cos (40° - 70°) + cos (40° + 70°) + 1/2 cos 70°]

⇒ √3/4 × [1/2 × (cos -30° + cos 110°) + 1/2 cos 70°]

We know,

cos (-θ) = cos θ

⇒ √3/4 × [1/2 × (cos 30° + cos 110°) + 1/2 cos 70°]

Substitute the value of cos 30°

⇒ √3/4 ( 1/2 × (√3/2  + cos 110°) + 1/2 cos 70°)

⇒ √3/4 (√3/4 + 1/2 × cos 110° + 1/2 cos 70°)

Now,

cos 110° = cos (180° - 70°)

Here cos lies in the second quadrant and is negative

cos 110° = -cos 70°

Substitute the value above,

⇒ √3/4 ( √3/4 + 1/2 × - cos 70° + 1/2 × cos 70°)

⇒ √3/4 ( √3/4 + 1/2 × cos 70° - 1/2 cos 70°)

⇒ √3/4 × √3/4

⇒ 3/16

= RHS

Hence proved.

Answer 2

Answer:

$\huge\rm{\underline{\underline{Solution:–}}}$

$\rm \: → \frac{1}{2} ×2(cos \: 10°. \: cos \: 50°).cos \: 30. \: cos \: 70 \\ \rm \: → \frac{1}{2} ×[cos(10+50)+cos(10-50)]. \: cos \: 30°$

$\rm \: →\frac{1}{2} ×[cos \: 60+cos \: (-40)].cos \: 30°.cos \: 70 \\ \rm \: → \frac{1}{2} ×[\frac{1}{2} +cos \: 40°].cos \: 30°-cos \: 70°$

$\rm \: → \frac{1}{2} ×[ \frac{1}{2} +cos \: 40°]. \frac{√3}{2} .cos \: 70° \\ \rm \: → \frac{√3}{4} \: cos \: 70°[ \frac{1}{2} +cos \: 40°]$

$\rm \: → \frac{√3}{8} cos \: 70°+ \frac{√3}{4} cos \: 40°.cos \: 70°$

$\rm \: → \frac{√3}{8} cos \: 70°+ \frac{√3}{4} × \frac{1}{2} ×[2 \: cos \: 40°.cos \: 70°]$

$\rm \: → \frac{√3}{8} \: cos \: 70°+ \frac{√3}{8} ×[cos (\: 40+70)]$

$\rm \: → \frac{√3}{8} \: cos \: 70°+ \frac{√3}{8} ×[cos \: 110+cos \: 30]$

$\rm \: → \frac{√3}{8} \: cos \: 70°+ \frac{√3}{8} ×[cos(180-70)+cos \: 30]$

$\rm \: → \frac{√-3}{8} \: cos \: 70°+ \frac{√3}{8} × \frac{√3}{2}$

$\rm \: Ans= \frac{3}{16}$