Math, asked by NarayanChaudhari, 8 months ago

cos(105°)+sin(105°)=cos45°​

Answers

Answered by Legend42
4

Answer:

Taking LHS:

sin 105° + cos 105°

cos (90° - 105°) + cos 105°

cos 15° + cos 105°

cos C + cos D = 2(cos(C+D)/2)(cos(C-D)/2)

2cos(120°/2)cos(90°/2)

2cos60°cos45°

2(1/2)cos45°

cos45°

= RHS

I could also have converted to sin:

sin 105° + sin (90° – 105°)

sin 105° - sin 15°

sin C - sin D = 2(cos(C+D)/2)(sin(C-D)/2)

2cos(120°/2)sin(90°/2)

2cos60°sin45°

2(1/2)sin45°

sin 45°

= cos 45°

= RHS

  • Route 2: Trigonometric Identities

We have:

sin 105° + cos 105° = cos 45°

Let me multiply both sides by 1/√2

Why? You'll see shortly.

(1/√2)(sin 105° + cos 105°) = (1/√2)cos 45°

You must know that (1/√2) = cos 45° = sin 45°

That gives me

(1/√2)sin 105° + (1/√2)cos 105° = (1/√2)cos45°

At this point, I can do two simple things:

  • Write that as:

sin45°sin105° + cos45°cos105° = cos^2 45°

sinAcosB + cosAsinB = sin(A+B)

cos (105° - 45°) = cos^2 45°

cos 60° = cos^2 45°

1/2 = 1/2

  • Write that as:

cos45°sin105° + sin45°cos105° = sin45°cos45°

2sinAcosA = sin(A+A) || sinAcosA = (1/2)sin(A+A)

sin(105° + 45°) = (1/2)sin(45° + 45°)

sin 150° = (1/2)sin 90°

sin (180° - 30°) = (1/2)

sin 30° = 1/2

1/2 = 1/2

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