Solving the Puzzle: 2 tan2A+sin2A-1

[1/2"]

Homework Statement

If A is an acute angle and sinA = cosA
Find the value of 2 tan² A+ sin² A-1

Homework Equations

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!

 

[Mark44]

Homework Statement

If A is an acute angle and sinA = cosA
Find the value of 2 tan² A+ sin² A-1

Homework Equations

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!

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).

 

[1/2"]

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).

But how do I use it?
Any clue ?

 

[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

 

[Mark44]

But how do I use it?
Any clue ?

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.

 

[Dickfore]

<br /> \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<br />

Substitute these in the expression:

<br /> 2 \tan^{2}{\alpha} + \sin^{2}{\alpha} - 1<br />

and do the arithmetic and you are done.

 

[Mark44]

<br /> \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<br />

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

Substitute these in the expression:

<br /> 2 \tan^{2}{\alpha} + \sin^{2}{\alpha} - 1<br />

and do the arithmetic and you are done.

 

[hunt_mat]

Or use this:
<br /> 1+\frac{\cos^{2}x}{\sin^{2}x}=\frac{1}{\sin^{2}x}<br />