Video: Knowing the Cosine, Sine, or Tangent of Two Angles and Finding the Cosine, Sine, or Tangent of Their Sum or Difference

Given that cos πœƒ = βˆ’βˆš(5)/3, where 0 ≀ πœƒ ≀ πœ‹, and cos πœ‘ = √(2)/3, where 0 ≀ πœ‘ ≀ πœ‹, find the exact value of cos (πœ‘ + πœƒ).

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Video Transcript

Given that cos πœƒ is equal to negative root five over three, where πœƒ is between zero and πœ‹, and cos πœ‘ is equal to root two over three, where πœ‘ is between zero and πœ‹, find the exact value of cos πœ‘ plus πœƒ.

Our goal is to find the value of cos πœ‘ plus πœƒ. You might be tempted to use the inverse cosine function or arc cosine function to find the values of πœƒ and πœ‘ and then add these two values and find the cosine of the sum. But we’re looking for an exact value, which your calculator may not give you if you use this method.

A better method is to use the multiple angle formula, which gives the cosine of a sum of two angles in terms of the cosine and sine of those two angles. We want to find cos πœ‘ plus πœƒ, so we can set 𝐴 equal to πœ‘ and 𝐡 equal to πœƒ. And it’s looking hopeful because we’re given the values of cos πœ‘ and cos πœƒ in the question. However, we’re not explicitly told the values of either sin πœ‘ or sin πœƒ. We’re going to have to work them out.

Let’s start by finding sin πœ‘. Again, we might be tempted to use our calculators to find the value of πœ‘. That would be inverse cosine of root two over three. And then having πœ‘, we could then use our calculators again to find the sin of πœ‘.

However, we’re looking for an exact value at the end, and there’s no guarantee that our calculator will give us the exact value. Instead, we’ll use the fact that cos of πœ‘ squared plus sin of πœ‘ squared is equal to one. This is true for any angle, and so it’s certainly true for πœ‘.

We know that cos πœ‘ is root two over three, and so cos πœ‘ squared is root two over three squared. And root two over three squared is two over nine. Rearranging, we find that sin of πœ‘ squared is seven over nine, and so sin πœ‘ is equal to plus or minus root seven over nine.

We have two possible values for sin πœ‘, and we’re going to have to decide between them. Otherwise, we’re going to end up with multiple possible values of cos πœ‘ plus πœƒ. We do this by using our extra bit of information that πœ‘ is between zero and πœ‹. And either by thinking about the unit circle or the graph of 𝑦 equals sin π‘₯, we can see that the sine function is positive on this interval from zero to πœ‹ radians.

Sin πœ‘ is therefore positive, and so it must be root seven over nine and not negative root seven over nine. And we know that we can rewrite this as root seven over three.

So now that we found the value of sin πœ‘, we can move on to finding the value of sin πœƒ. And we find this value in exactly the same way as we found the value of sin πœ‘. We write down the relationship between sin πœƒ and cos πœƒ, substitute in the value of cos πœƒ that we have from the question, and do some algebra to find that sin πœƒ is either two over three or negative two over three. And just as before, because πœƒ is between zero and πœ‹, sin πœƒ will be positive or at least nonnegative. And so sin πœƒ must be two over three and not negative two over three.

Having found the value of sin πœƒ, we now have the values of cos πœ‘ and cos πœƒ from the question and sin πœ‘ and sin πœƒ which we’ve worked out. And so we’re ready to substitute into our formula. Cos πœ‘ is root two over three, cos πœƒ is negative root five over three, sin πœ‘ is root seven over three, and sin πœƒ is two over three. The rest is just arithmetic.

We find the two products using the fact that root two times root five is root 10, to get negative root 10 over nine minus two root seven over nine. And swapping the order of these two terms and then combining them into one fraction using the fact that they have a common denominator, we get negative two root seven plus root 10 over nine.

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