Triogonoeasy

Geometry Level 1

sin ⁡ 7 5 ∘ − cos ⁡ 1 5 ∘ = ? \large \sin 75^{\circ} - \cos 15^{\circ}= \ ?


The answer is 0.

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Math Man by math man , 20, Indonesia

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4 solutions

Sandeep Bhardwaj
Sep 18, 2014

s i n X = c o s ( 9 0 0 − X ) sinX=cos(90^0-X)

so s i n 7 5 0 = c o s 1 5 0 sin75^0=cos15^0

s i n 7 5 0 − c o s 1 5 0 = c o s 1 5 0 − c o s 1 5 0 = 0 sin75^0-cos15^0=cos15^0-cos15^0=\boxed{0}

8 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

Yes, can you find the exact form of sin ⁡ ( 7 5 ∘ ) \sin(75^\circ) instead?

Vikram Venkat
Apr 3, 2015

Since s i n sin and c o s cos values for any two complimentary angles are the same, it equates to 0 0

3 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

Nice. This is the solution intended.

Vampy Vampy
Apr 3, 2015

Sin75° - cos15° = sin75° - sin(90-15) because cos x = sin (90-x) = sin75° - sin75° = 0

1 Helpful 0 Interesting 0 Brilliant 0 Confused
Math Man
Sep 18, 2014

sin(x)=cos(90-x)

0 Helpful 0 Interesting 0 Brilliant 0 Confused

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