if you were to find the cos75°, which formulas could you use? Would you get the same answer using both methods? How would you know?

please explain

if you were to find the cos75°, which formulas could you use? Would you get the same answer using both methods? How would you know?

please explain

Tutors, please sign in to answer this question.

Brookline, MA

You can use the sum of angles formula.

cos 75° = cos 30° cos 45° - sin 30° sin 45° = (√3/2)(√2/2) - (1/2)(√2/2) = (√6 - √2)/4

You can also use the half-angle formula.

cos 75° = √[(1+cos 150°)/2] = √(2-√3) / 2

Since both answers are positive, you can verify that these two are equal by squaring them.

[(√6 - √2)/4]^{2} = (6+2-2·√6·√2)/16 = (2-√3)/4

{√(2-√3) / 2}^{2} = (2-√3)/4

Wilton, CT

No matter how you found it, the answer would be the same.

You want to know how in terms of sines and cosines you already know and addition formulas.

Let's do it. One way 75°=90°-15°=90°-1/2(30°)

cos(x-y)=cosxcosy+sinxsiny

we need COS15°=cos(1/2(30°))=√((1+cos30°)/2)=√((1+√3/2)/2)

SIN15°=√((1-cos30°)/2)=√((1-√3/2)/2)

NOW cos(90°-15°)=cos90°cos15°+sin90°sin15°

COS75°=0+1*√((1-√3/2)/2)

OR BY KNOWING THAT CO-FUNCTIONS OF COMPLEMENTARY ANGLES ARE EQUAL

COS75°=SIN15° AND USING Sin15°=sin(1/2(30°))=√((1-cos30°)/2)=√((1-√3/2)/2

- Algebra 2 3503
- Trigonometry 1526
- Trigonometric Functions 220
- Math 9857
- Math Help 5445
- Algebra 5057
- Math Word Problem 4500
- Calculus 2205
- Precalculus 1577
- Word Problem Help 1163

## Comments