Homework Help: Trig sum

1. Jul 19, 2010

1/2"

1. The problem statement, all variables and given/known data
If A is an acute angle and sinA = cosA
Find the value of 2 tan2 A+ sin2 A-1

2. Relevant equations

3. The attempt at a solution
I well tried this way >>
2 (sin 2 A/cos 2A) +sin 2 A-1
= 1+ cos 2 A
But I am unable to get the 'value' the answer given in the book is 3/2.
Can anybody please suggest any way by which I can get a proper value ?
I would be very thankful!

2. Jul 19, 2010

Staff: Mentor

You aren't using all of the given information. The piece you are not using is that sin(A) = cos(A). Since A is acute (less than 90 deg.), there is only one angle for which sin(A) = cos(A).

3. Jul 19, 2010

1/2"

But how do I use it?
Any clue ?

Last edited: Jul 19, 2010
4. Jul 19, 2010

eumyang

If you really don't know, you could make a table in a spreadsheet or something. Put A in one column, sin A in the 2nd column, and cos A in the 3rd, and put some values in for A to see where sin A = cos A.

Or, you have a graphing calculator, graph Y1=sin(x) and Y2=cos(x) and see where they intersect. Change the window so that the x-values only go from 0 to pi/2 (you have to be in radians here).

Or, use one of the cofunction identities.

69

5. Jul 19, 2010

Staff: Mentor

In addition to what eumyang said, there are some first quadrant angles for which you should know the exact value of the sine, cosine, and tangent functions. These angles are 0, pi/6, pi/4, pi/3, and pi/2.

6. Jul 19, 2010

Dickfore

$$\sin{\alpha} = \cos{\alpha} \Rightarrow 1 = \sin^{2}{\alpha} + \cos^{2}{\alpha} = 2 \, \sin^{2}{\alpha} \; \wedge \; \tan{\alpha} = \frac{\sin{\alpha}}{\cos{\alpha}} = 1$$

Substitute these in the expression:

$$2 \tan^{2}{\alpha} + \sin^{2}{\alpha} - 1$$

and do the arithmetic and you are done.

7. Jul 19, 2010

Staff: Mentor

This is sort of the long way around to get there.
$$sin(A) = cos(A) \Rightarrow \frac{sin(A)}{sin(A)} = 1 \Rightarrow tan(A) = 1$$
for $A \neq \pi/2 + k \pi$

8. Jul 20, 2010

hunt_mat

Or use this:
$$1+\frac{\cos^{2}x}{\sin^{2}x}=\frac{1}{\sin^{2}x}$$

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