another pre-calculus problem i need help with?

Question

find the exact value of
cos(pi/16)cos(3pi/16) - sin(pi/16)sin(3pi/16)

I figured i had to use cos(u+v)=cosUcosv - sinUsinv but i just got confused... help!

Answer 1

You are correct.

cos(u + v) = cos(u)cos(v) - sin(u)sin(v)

In our case, u = &pi/16, and v = 3&pi/16, so

cos(u + v) = cos(&pi/16 + 3&pi/16)
= cos(4&pi/16)
= cos(&pi/4)
= √(2)/2

Answer 2

You are on exactly the right track -- this formula fits the pattern of the question.

U = pi/16
V = 3pi/16

Substituting: cos(U+V) = cos( pi/16 + 3pi/16).

Answer 3

Yes, you are definitely on the right track! If you use that formula backwards, you have cos (π/16 + 3π/16) = cos (4π/16) = cos (π/4) = √2 / 2

Answer 4

The x intercepts are the place the graph touches the x axis at those factors the value of y is 0 so in basic terms set the equation equivalent to 0 remedy for x and you're transforming into your x intercepts -x^5+2x^3=0 remedy for x Factoring technique making use of the suitable consumer-friendly denominator -x^3(x^2-2) through fact the equation equals 0 we can deduce that the two the parenthesis is comparable to 0 or the term multiplying the parentjesis is 0 so set them the two equivalent to 0 and remedy for x -x^3=0 x=0 And x^2-2=0 factor (x-?2)(x+?2) x=0,?2,-?2

Answer 5

what you have is cos(π/16 + 3π/16) = cos(4π/16) = cos(π/4) = √2 / 2.