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Solving Trig Problems

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Date: 08/18/97 at 04:44:07
From: Venessa
Subject: Solving trigonometry

I'm studying in high school and got these problems from past year
questions:

1. Find the value of Tan [(x+1)/(x-1)]^(-1) + Tan (x)^(-1).

2. Given that sin[x]^2 + 6 sin[x]cos[x] + 9 cos[x]^2 = A + B sin[2x] +
C cos[2x], find the values of A, B and C.

Hence, or otherwise, find the maximum and minimum values of
sin[x]^2 + 6 sin[x]cos[x] + 9 cos[x]^2.

I have tried all the related formulas in many ways for question 1, but
I still can't solve this problem. For question 2, I don't even know
where to start. Your help would be really appreciated.

Venessa
```

```
Date: 08/22/97 at 15:57:02
From: Doctor Rob
Subject: Re: Solving trigonometry

1. I am assuming that Tan[a]^(-1) means Arctan[a]. If so, it may be
hard to find the angle, but I can find its tangent:

Tan[Arctan[(x+1)/(x-1)] + Arctan[x]]
= [(x+1)/(x-1) + x]/[1 - x*(x+1)/(x-1)],

using the fact that Tan[Arctan[y]] = y and the formula for the
tangent of the sum of two angles,

Tan[u+v] = (Tan[u]+Tan[v])/(1-Tan[u]*Tan[v])

and then this

= [x+1+x*(x-1)]/[x-1-x*(x+1)]
= (x^2+1)/(-x^2-1)
= -1.

This means that the angle is one whose tangent is -1, or 3*Pi/2.

Neat problem!

2. There are at least two ways to find these A, B, and C.

First, plug in three different values for x, giving three equations
in three unknowns:

x = 0:  9 = A + C
x = Pi/4:  8 = A + B
x = -Pi/4:  2 = A - B

Solving together, A = 5, B = 3, and C = 4.

Second, use double-angle formulas:

sin^2[x] = (1 - cos[2*x])/2
cos^2[x] = (1 + cos[2*x])/2
sin[x]*cos[x] = sin[2*x]/2

Then

sin^2[x] + 6*sin[x]*cos[x] + 9*cos^2[x]
= 1/2 - cos[2*x]/2 + 3*sin[2*x] + 9/2 + 9*cos[2*x]/2
= 5 + 3*sin[2*x] + 4*cos[2*x]

To continue with the second part, this

= 5 + 5*[(3/5)*sin[2*x] + (4/5)*cos[2*x]]
= 5 + 5*sin[2*x+a]

where cos[a] = 3/5 and sin[a] = 4/5. Since the sine has maximum
value 1, 5 + 5*sin[2*x+a] has maximum value 5 + 5*1 = 10.

Otherwise,

sin^2[x] + 6*sin[x]*cos[x] + 9*cos^2[x]
= (sin[x] + 3*cos[x])^2
= 10*(sin[x]/Sqrt[10] + cos[x]*3/Sqrt[10])^2
since 10 = 1^2 + 3^2, so
1 = (1/Sqrt[10])^2 + (3/Sqrt[10])^2),
= 10*sin[x+b]^2,
where b is an angle such that
cos[b] = 1/Sqrt[10] and sin[b] = 3/Sqrt[10].

Since the maximum absolute value of the sine is 1, the maximum value
of the square of the sine is 1, and the maximum value of this function
is 10.

This is an old trick to simplify(?) u*sin[x] + v*cos[x] by converting
it to the form Sqrt[u^2+v^2]*sin[x+b], where cos[b] = u/Sqrt[u^2+v^2].
I used this twice above.

-Doctor Rob,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```
Associated Topics:
High School Trigonometry

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