Video: Simplifying Trigonometric Expressions Using Double-Angle Identities

Simplify sin (π‘₯ + 𝑦) cos 𝑦 βˆ’ cos (π‘₯ + 𝑦) sin 𝑦.

02:43

Video Transcript

Simplify sin π‘₯ plus 𝑦 times cos 𝑦 minus cos π‘₯ plus 𝑦 times sin 𝑦.

To simplify this, we use the angle sum identities for sin and cos which allow us to write that sin π‘₯ plus 𝑦 and cos π‘₯ plus 𝑦 in terms of sin π‘₯, sin 𝑦, cos π‘₯, and cos 𝑦. We write out the expression we needed to simplify again and reapply one of the angles sum identities to rewrite sin π‘₯ plus 𝑦, like so. And of course, we need to multiply this by cos 𝑦 before subtracting the other term.

Now we can shift our focus to the other term. This term involves cos π‘₯ plus 𝑦 which we can rewrite using the other identity. And we shouldn’t forget that we need to multiply this by sin 𝑦. Having applied the identities, we just need to expand out the parentheses. The first term we get is sin π‘₯ cos squared 𝑦. To that, we add cos π‘₯ sin 𝑦 cos 𝑦. Expanding the other parentheses, and remembering that we need to subtract these terms, we get minus cos π‘₯ cos 𝑦 sin 𝑦. And the two minuses make a plus sin π‘₯ sin squared 𝑦. We notice that the two terms in the middle are identical apart from the order of their factors, and so they cancel. And the two terms that we’re left with have a common factor of sin π‘₯ which we’ll take out. Taking this factor out, we have sin π‘₯ times cos squared 𝑦 plus sin squared 𝑦 and cos squared something plus sin squared something, which is obviously the same as sin squared something plus cos squared something, is equal to one.

Hence, our final answer is sin π‘₯. This is a more straightforward way of answering the question, but there is another fun method. You might know that there’s a formula for the sine of a difference of two angles, sin 𝐴 minus 𝐡. If you don’t know it, then you can derive it from the angle sum identity we have on screen, setting π‘₯ equals 𝐴 and 𝑦 equals negative 𝐡. And further using the facts that cos of minus 𝐡 equals cos of 𝐡 and sin of minus 𝐡 equals minus sin of 𝐡.

Having derived it, we can compare it with the expression we have to simplify and notice that it’s very similar. In fact, this expression that we have to simplify is exactly the right-hand side of this identity, setting 𝐴 equal to π‘₯ plus 𝑦 and 𝐡 equal to 𝑦. Substituting the values of 𝐴 and 𝐡 in the left-hand side of our identity and simplifying, we get sin π‘₯.

So this was a much quicker method for simplifying the expression. However, it did require us to recognize a slightly disguised angle difference identity.

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