Prove that cos x + cos 3x + cos 5x + cos 7x = 4cos x cos 2x cos 4x

Step-by-step explanation:

Prove the following?

cos7x+cos5x+cos3x+cosx=4cosxcos2xcos4x

2

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Answer

3.4/5

36

LHS=cos7x + cos5x + cos3x + cosx

=2cos[(7x+5x)/2].cos[(7x-5x)/2] + 2cos[(3x+x)/2].cos[(3x-x)/2]

[∵ cosA + cosB = 2.cos[(A+B)/2].cos[(A-B)/2]]

=2cos[(12x)/2].cos[(2x)/2] + 2cos[(4x)/2].cos[(2x)/2]

=2cos 6x.cosx+2cos2x.cosx

=2cosx(cos 6x+ cos 2x)

=2.cosx(2cos[(6x+2x)/2].cos[(6x-2x)/2]

=2.cosx(2cos4x.cos2x)

=4.cosx.cos2x.cos4x

=RHS.

hope this helps you :

80 the answer is 80

step-by-step explanation:

answer:   the paragraph proof is mentioned below.

step-by-step explanation:

given: in a δxwy where

such that, ∠wzx ≅ ∠yzx and zw≅zy

we have to prove that: segment zx is a perpendicular bisector of segment wy.

prove: since, zw≅zy ⇒ ∠wzy is a supplementary angle.

then, ∠wzy= 180°

⇒∠wzx + ∠yzx = 180° ( because xz divides ∠wzy into two parts)

but, ∠wzx ≅ ∠yzx ( given)

⇒∠wzy+∠wzy=180° ( by substitution)

⇒2∠wzy=180°

⇒∠wzy=90°

⇒∠wzy=∠yzx =90°

therefore, xz⊥wy where wz=zy

⇒zx is a perpendicular bisector of wy.