Question Video: Using Double Angle Identities to Evaluate Trigonometric Expressions | Nagwa Question Video: Using Double Angle Identities to Evaluate Trigonometric Expressions | Nagwa

Question Video: Using Double Angle Identities to Evaluate Trigonometric Expressions Mathematics

Find, without using a calculator, (1 βˆ’ cos 2𝑋)/(1 + cos 2𝑋) given tan 𝑋 = 4, where 𝑋 ∈ (πœ‹, 3πœ‹/2).

02:16

Video Transcript

Find, without using a calculator, one minus the cos of two 𝑋 over one plus the cos of two 𝑋 given tan of 𝑋 equals four, where 𝑋 is an element of the interval πœ‹, three πœ‹ over two.

Alright, so this is the fraction we want to solve for. And to help us on our way, we’re told that the tan of 𝑋 equals four and that 𝑋, which is an angle, exists somewhere in the third quadrant, from πœ‹ to three πœ‹ over two radians. Looking at the expression we want to solve for, the first thing we can notice is that we’re taking the cos of two times this angle 𝑋. This suggests that we can apply what’s called a double-angle identity.

And for the cosine function, there are three different forms of this identity. With our aim being to simplify this given fraction as much as possible, we can choose the particular forms of the cosine’s double-angle identity that enable that. For example, in the numerator, notice that we have one minus the cos of two 𝑋. If we were to replace the cos of two 𝑋 by one minus two times the sin squared of 𝑋, then notice that the positive one and negative one in our numerator add up to zero. Similarly, if we replace the cos of two 𝑋 in our denominator with two times the cos squared of 𝑋 minus one, then here again we have a positive one and a negative one adding up to zero.

Accounting for all the sines involved, this gives us two times the sin squared of 𝑋 over two times the cos squared of 𝑋. And we can see from here that the twos will cancel from top and bottom. Noticing that we have a fraction of a power of a sine over a power of a cosine, we can recall at this point that the tan of an angle equals the sin of that angle over the cos of that angle. This implies that the tan squared of 𝑋 equals the sin squared of 𝑋 over the cos squared of 𝑋. And therefore, our original expression simplifies to the tan squared of 𝑋.

In our problem statement, we’re told that the tan of 𝑋 is four and also that this angle 𝑋 exists in the third quadrant. In this quadrant, the tangent function is always positive. And this means we can safely substitute in four for the tan of 𝑋 and then square it to get a result of positive 16. This is the value of one minus the cos of two 𝑋 divided by one plus the cos of two 𝑋.

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