If cos 9 alpha = sin alpha 9 alpha < 90 degree then 5 alpha =
what does "sin alpha 9" mean?
To solve this problem, we need to use the trigonometric identity for the cosine of a sum of angles:
cos(A + B) = cos A cos B - sin A sin B
Let's substitute A = 9alpha and B = alpha into the above formula:
cos(9alpha + alpha) = cos(9alpha) cos(alpha) - sin(9alpha) sin(alpha)
Simplifying further:
cos(10alpha) = cos(9alpha) cos(alpha) - sin(9alpha) sin(alpha)
Now, we can apply the given condition that cos(9alpha) = sin(alpha), which means we can substitute cos(9alpha) with sin(alpha):
cos(10alpha) = sin(alpha) cos(alpha) - sin(9alpha) sin(alpha)
Next, let's use the Pythagorean identity sin²(x) + cos²(x) = 1 to substitute sin²(alpha) for 1 - cos²(alpha):
cos(10alpha) = sin(alpha) cos(alpha) - sin(9alpha) sin(alpha)
cos(10alpha) = sin(alpha) cos(alpha) - sin(9alpha) (1 - cos²(alpha))
Expanding further:
cos(10alpha) = sin(alpha) cos(alpha) - sin(9alpha) + sin(9alpha) cos²(alpha)
Now, we can factor out sin(9alpha) from the last two terms:
cos(10alpha) = sin(alpha) cos(alpha) + sin(9alpha) (cos²(alpha) - 1)
Using the Pythagorean identity cos²(x) = 1 - sin²(x), we can simplify cos²(alpha) - 1 to -sin²(alpha):
cos(10alpha) = sin(alpha) cos(alpha) + sin(9alpha) (-sin²(alpha))
cos(10alpha) = sin(alpha) cos(alpha) - sin(9alpha) sin²(alpha)
Finally, we can substitute cos(9alpha) = sin(alpha) back into the equation:
cos(10alpha) = cos(9alpha) cos(alpha) - sin(9alpha) sin(alpha)
cos(10alpha) = cos(9alpha + alpha) = cos(10alpha)
Since cos(10alpha) is equal to itself, we can conclude that the value of 5alpha doesn't affect the equation. Therefore, 5alpha can be any value.