Video: AQA GCSE Mathematics Higher Tier Pack 5 • Paper 1 • Question 19

Write 3 sin (45°) + 7 cos (45°) − 6 tan (45°) in the form √(𝑎) + 𝑏, where 𝑎 and 𝑏 are integers.

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Video Transcript

Write three sin 45 degrees plus seven cos 45 degrees minus six tan 45 degrees in the form root 𝑎 plus 𝑏, where 𝑎 and 𝑏 are integers.

There are several trigonometric values we need to know by heart and there’s a little trick we can use to help us. We use this table and it will help us work out the values for sin 𝜃, cos 𝜃, and tan 𝜃 when 𝜃 is 30 degrees, 45 degrees, and 60 degrees. We begin by writing one, two, three; three, two, one in the first two rows. We turn each of these numbers into a fraction by adding a denominator of two. We then find the square root of the numerator in each case. Remember though the square root of one is simply one. So we don’t worry about writing it on sin of a 30 and cos of 60.

There are no nice little tricks to help us remember the values for tan 𝜃. We’re just going to need to learn these. We can use this table now to work out the values of sin of 45, cos of 45, and tan of 45 degrees as required in the question.

Sin of 45 degrees we can see is root two over two. So three sin of 45 degrees is three multiplied by root two over two. And if we treat three as a fraction with a denominator of one, we can simply multiply the numerators, three multiplied by root two is three root two, and then the denominators, one multiplied by two is two.

We repeat this process with seven cos of 45 degrees. Cos 45 is root two over two. So we multiply this by seven. And once again, treating seven like a fraction with the denominator of one, we get seven root two over two. Tan of 45 is a little easier. It’s one. So six tan 45 is six multiplied by one which is of course six.

And we can now say that three sin of 45 degrees plus seven cos of 45 degrees minus six tan of 45 degrees is three root two over two plus seven root two over two minus six. The denominators of these two fractions is the same. So we can simply add their numerators. Three root two plus seven root two becomes 10 root two. So our expression becomes 10 root two over two minus six and 10 divided by two is five. So we have five root two minus six.

We aren’t finished though. We were told to write our answer in the form root 𝑎 plus 𝑏, where 𝑎 and 𝑏 are integers. That’s whole numbers. Our value for 𝑏 is negative six. That is indeed a whole number, so we’re done there. But we have five root two rather than a single root of some whole number. We’re going to need to manipulate five root two and we’re going to use this fact: the square root of 𝑎 multiplied by the square root of 𝑏 is the square root of 𝑎 multiplied by 𝑏.

And since we know that five is the square root of 25, we’re going to rewrite this a little bit as the square root of 25 multiplied by the square root of two. 25 multiplied by two is 50. So using that rule we just mentioned, the square root of 25 multiplied by the square root of two is the square root of 50. And we’re finished.

Three sin of 45 degrees plus seven cos of 45 degrees minus six tan of 45 degrees can be written as root 50 minus six. And we weren’t asked to specify our values for 𝑎 and 𝑏. But we did already say that 𝑏 was equal to negative six. So for completeness, we can say that 𝑎 is equal to 50, which is of course an integer, a whole number. We’re done.

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