# Math Help - Trigonometry surds

1. ## Trigonometry surds

If $sinA=\frac{2}{\sqrt{5}}$ and $tanB=\frac{1}{2}$, how can I prove that $A-B=45$?

Using a calculator to find inverse sine and tangent then subtracting the angles gets 45 degrees, but I do not think that is the correct way to approach the question.

Similar questions are $sin(A+B+C)=\frac{2\sqrt{2}}{3}$ find $sinC$ etc.

I think when I get the method I should be fine in doing these questions. Thanks!

2. Originally Posted by RAz
If $sinA=\frac{2}{\sqrt{5}}$ and $tanB=\frac{1}{2}$, how can I prove that $A-B=45$?

Using a calculator to find inverse sine and tangent then subtracting the angles gets 45 degrees, but I do not think that is the correct way to approach the question.

Similar questions are $sin(A+B+C)=\frac{2\sqrt{2}}{3}$ find $sinC$ etc.

I think when I get the method I should be fine in doing these questions. Thanks!
$sinA = \frac{2}{\sqrt{5}}$

$CosA = \sqrt{1-sin^2(A)}$

Then find tanA.

Now $tan(A-B) = \frac{tanA - tanB}{1+tanAtanB}$

Substitute the values of tanA and tanB and simplify.

3. You might try using simple sketches.

$sin A = \frac{2}{\sqrt{5}}$

Then, draw a simple right angled triangle, with hypotenuse [/tex]\sqrt5[/tex], opposite side 2 and adjacent side 1.

Ah, coincidence, the angle opposite to the angle A is the angle B!

Now that you know all the trigonometric ratios of A and B, you can use any identity you know;

$sin(A-B) = sinAcosA - sinBcosB$

$cos(A-B) = cosAcosB + sinAsinB$

or the tangent ratio given by sa-ri-ga-ma.

For cos and sin, you should get $cos(A-B) = sin(A-B) = \frac{1}{\sqrt2}$

For tan, you should get tan(A-B) = 1.

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For the second question, provided you know A, B, or one trigonometric identity of A and B, you use the identity:

$sin(A+B) = sinAcosA + sinBcosB$

Where, you replace A by (A+B) and B by C.

Then, you can use the identity : $sin(2A) = 2 sinA cos A$ to solve for C.