Find, without using a calculator, the value of sin two 𝐵 divided by two multiplied by cos two 𝐵 given that the cos of angle 𝐵 is equal to four-fifths, where 𝐵 is greater than three 𝜋 over two and less than two 𝜋.
Before focusing on the expression we need to calculate, let’s look at the information we have been given. We are told that the cos of angle 𝐵 is equal to four-fifths and that this angle 𝐵 lies between three 𝜋 over two and two 𝜋 radians. The CAST diagram shown helps us identify whether the sine, cosine, and tangent of any angle between zero and two 𝜋 is positive or negative. In this question, we know that the angle 𝐵 lies in the fourth quadrant. In this quadrant, the cosine of the angle is positive, whereas both the sine of the angle and tangent of the angle are negative. We are also told in the question that the cos of angle 𝐵 is equal to four-fifths. We can use this information together with our trigonometrical identities to calculate the sin of angle 𝐵 and the tan of angle 𝐵.
We recall that sin squared 𝜃 plus cos squared 𝜃 is equal to one. This means that sin squared 𝐵 plus four-fifths squared is equal to one. We can square four-fifths by squaring the numerator and denominator separately, giving us 16 over 25. We can then subtract this from both sides of our equation. sin squared 𝐵 is therefore equal to nine over 25. We can then square root both sides of this equation such that the sin of angle 𝐵 is equal to the positive or negative square root of nine over 25. This simplifies to positive or negative three-fifths. Recalling that the sin of angle 𝐵 must be negative, this is equal to negative three-fifths.
We know that the tan of any angle 𝜃 is equal to the sine of that angle divided by the cosine of the angle. This means that the tan of angle 𝐵 is equal to negative three-fifths divided by four-fifths. The right-hand side simplifies to negative three-quarters. We now have values for the sin of angle 𝐵, the cos of angle 𝐵, and the tan of angle 𝐵. Our next step is to try to substitute these into the expression sin of two 𝐵 divided by two multiplied by the cos of two 𝐵. We can do this using our knowledge of the double-angle formulae. The sin of two 𝜃 is equal to two multiplied by the sin of 𝜃 multiplied by the cos of 𝜃. And the cos of two 𝜃 is equal to cos squared 𝜃 minus sin squared 𝜃.
The expression sin two 𝐵 over two multiplied by cos two 𝐵 can be rewritten as shown. Dividing the numerator and denominator by two, we are left with sin 𝐵 multiplied by cos 𝐵 all divided by cos squared 𝐵 minus sin squared 𝐵. We can now substitute the values negative three-fifths and four-fifths into this expression. The numerator becomes negative three-fifths multiplied by four-fifths, and the denominator is four-fifths squared minus negative three-fifths squared. Negative three-fifths multiplied by four-fifths is equal to negative twelve twenty-fifths. And four-fifths squared minus negative three-fifths squared is equal to seven twenty-fifths. This simplifies to negative 12 over seven. The value of sin two 𝐵 divided by two multiplied by the cos of two 𝐵 is negative twelve-sevenths.
An alternative method here would be to recognize that sin two 𝐵 divided by cos two 𝐵 is equal to tan of two 𝐵. Our expression could therefore be rewritten as tan of two 𝐵 divided by two. We could then have used the fact that the tan of two 𝜃 is equal to two multiplied by the tan of 𝜃 divided by one minus tan squared 𝜃. Our expression would therefore have simplified to the tan of 𝐵 divided by one minus tan squared 𝐵. Substituting negative three-quarters for the tan of 𝐵 would once again have given us an answer of negative twelve-sevenths.