Question

how do i find the exact value under the given conditions.

sin alpha=20/29, 0 12 years ago

Answer 1

can anybody help with this????

Answer 2

\(\bf cos(\alpha+\beta)=cos(\alpha)cos(\beta)-sin(\alpha)sin(\beta)\\ \quad \\
sin(\beta)=\cfrac{20}{29}\implies \cfrac{opposite}{hypotenuse}\implies \cfrac{b}{c}\\ \quad \\
c^2=a^2+b^2\implies \sqrt{c^2-b^2}=a\qquad cos(\alpha)=\cfrac{a}{c}\\ \quad \\ \quad \\
-----------------------\\
cos(\beta)=\cfrac{12}{13}\implies \cfrac{adjacent}{hypotenuse}\implies \cfrac{a}{c}\\ \quad \\
c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b\qquad sin(\beta)=\cfrac{b}{c}\)

Answer 3

well... ahemm the sine above . should be alpha but anyhow \(\bf cos(\alpha+\beta)=cos(\alpha)cos(\beta)-sin(\alpha)sin(\beta)\\ \quad \\
sin(\alpha)=\cfrac{20}{29}\implies \cfrac{opposite}{hypotenuse}\implies \cfrac{b}{c}\\ \quad \\
c^2=a^2+b^2\implies \sqrt{c^2-b^2}=a\qquad cos(\alpha)=\cfrac{a}{c}\\ \quad \\ \quad \\
-----------------------\\
cos(\beta)=\cfrac{12}{13}\implies \cfrac{adjacent}{hypotenuse}\implies \cfrac{a}{c}\\ \quad \\
c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b\qquad sin(\beta)=\cfrac{b}{c}\)

you'd use the pythagorean theorem to find the missing "leg" or side and get the other trig function, then plug them in the cosine angle sum identity

Answer 4

got it i will try to figure that out

Answer 5

thanks i figured the right answer out