Trigonometry surds

**RAz** · June 7th 2010, 03:37 AM

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!

**sa-ri-ga-ma** · June 7th 2010, 05:45 AM

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.

**Unknown008** · June 7th 2010, 06:20 AM

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.

~~~~~~~~~

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.

Thread: Trigonometry surds

LinkBack

Thread Tools

Display

Trigonometry surds

Similar Math Help Forum Discussions

Trigonometry to Memorize, and Trigonometry to Derive

Trigonometry and Surds

surds

surds

surds

Search Tags