Using the complementary angle ratio results,find the value of x for the following. (a) cos(x+35)degree=sin x degree (b) sec x degree=cosec(x+20)degree
To find the value of \(x\) in each of the given trigonometric equations, we will use the complementary angle ratio identities.
(a) \( \cos(x+35) = \sin(x) \)
We'll start by rewriting the equation using the complementary angle ratio identity:
\( \cos(x+35) = \cos(90 - x) \)
Using the identity \( \cos(\theta) = \cos(-\theta) \), we can rewrite the equation as:
\( \cos(x+35) = \cos(x - 90) \)
Since the cosines of two angles are equal, the angles must be either the same or supplementary (their sum is 180 degrees).
So, we can either have:
\( x + 35 = x - 90 \) or \( x + 35 = 180 - (x - 90) \)
Solving the first equation, we find:
\( x - x = 35 + 90 \)
\( 0 = 125 \)
This is not possible, so we turn to the second equation:
\( x + 35 = 180 - x + 90 \)
Simplifying the equation:
\( 2x = 180 + 90 - 35 \)
\( 2x = 235 \)
Finally, solving for \( x \):
\( x = \frac{235}{2} \)
(b) \( \sec(x) = \csc(x + 20) \)
Again, we'll start by rewriting the equation using the complementary angle ratio identity:
\( \sec(x) = \csc(90 - (x + 20)) \)
Using the identity \( \sec(\theta) = \csc(90 - \theta) \), we have:
\( \sec(x) = \csc(70 - x) \)
Since these are reciprocal trigonometric functions, the angles must be either the same or supplementary.
So, we can either have:
\( x = 70 - x \) or \( x = 180 - (70 - x) \)
Solving the first equation gives:
\( 2x = 70 \)
\( x = 35 \)
This is a valid solution.
Solving the second equation:
\( x = 180 - 70 + x \)
Simplifying the equation:
\( 2x = 110 \)
\( x = 55 \)
This is another valid solution.
Therefore, the possible values of \( x \) for equation (b) are \( x = 35 \) and \( x = 55 \).