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# 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?

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?

### 2 Answers by Expert Tutors

Roman C. | Masters of Education Graduate with Mathematics ExpertiseMasters of Education Graduate with Mathe...
4.9 4.9 (316 lesson ratings) (316)
1
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

Hey Roman, I am so glad you're back! I like your brief and accurate answers.
Thank you.
Michael F. | Mathematics TutorMathematics Tutor
4.7 4.7 (6 lesson ratings) (6)
0
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