Kindly offer me help on trigonometry.

[brandon90]

Hi all,

Please help me solve the following trigonometric equations (1-4).Give all positive values of the angle between 0* and 360* that will satisfy each.Give any approximate value to the nearest minute only.

1.) cos 2x - sin^2 x/2 + 3/4 = 0

2.) cos 4x = sin 2x

3.) 3 sin (theta) - 4 cos (theta) = 2

4.) tan (x+15*) = 3 tan x

Verify
the identity:

1.) tan (2 tan^-1 x) = 2 tan (tan^-1 x+tan^-1 x^3)

Many thanks!

 

[morson]

For the first one, put 2u = x and multiply both sides by 4:

\(\displaystyle 4cos(4u) - 4sin^2(u) + 3 = 0\)

\(\displaystyle \rightarrow 4(1 - 2sin^2(2u)) - 4sin^2(u) + 3 = 0\)

\(\displaystyle \rightarrow 4 - 8sin^2(2u) - 4sin^2(u) + 3 = 0\)

\(\displaystyle \rightarrow 8sin^2(2u) + 4sin^2(u) - 7 = 0\)

Then applying the double angle formula again, we arrive at:

\(\displaystyle 32sin^4(u) - 36sin^2(u) + 7 = 0\)

\(\displaystyle (4sin^2(u) - 1)(8sin^2(u) - 7)\)

There must be some quicker way though that I'm just not seeing.

For the second one, put u = 2x:

\(\displaystyle cos(2u) = sin(u)\)

using \(\displaystyle cos(2u) = 1 - 2sin^2(u)\), so the equation becomes:

\(\displaystyle 2sin^2(u) + sin(u) - 1 = 0\)

and it's easy to see that:

\(\displaystyle (sin(u) + 1)(2sin(u) - 1) = 0\)

For the third one:

\(\displaystyle 3sin(x) - 4cos(x) = 2\)

It turns out the left side of the equation can be written \(\displaystyle 5sin(x - arcCos(\frac{3}{5}))\).

By the way, \(\displaystyle arcCos(\frac{3}{5}\)\) is the angle in the first quadrant.

This can be done by setting it equal to rsin(x - a), expanding it, and equating coefficients.

For the fourth one:

Use the addition formula for tan(a + b).

For the fifth one:

\(\displaystyle f(f^{-1}(x)) = x\) if they're both continuous at x.

I'd then use the identity above, and the double-angle identity, to simplify the left-hand side fully to \(\displaystyle \frac{2x}{1 - x^2}\\). I'd then simplify the right-hand side until you reach \(\displaystyle \frac{2x}{1 - x^2}\\) using the identity I posted a few lines up. Try expanding the right side using the formula for tan(a + b). You must also note that there are certainly values of x for which this particular identity is not true ( |x| = 1, for example)

 

[Deleted member 4993]

Hi all,

Please help me solve the following trigonometric equations (1-4).Give all positive values of the angle between 0* and 360* that will satisfy each.Give any approximate value to the nearest minute only.

1.) cos 2x - sin^2 x/2 + 3/4 = 0

2cos^2(x) - 1 + 1/2*{1 - 2 sin^2 (x/2)} + 1/4 = 0

2cos^2(x) + 1/2* cos(x) - 3/4 = 0

cos(x) = (1/2) , (-3/4) .....edited


2.) cos 4x = sin 2x

3.) 3 sin (theta) - 4 cos (theta) = 2

4.) tan (x+15*) = 3 tan x

Verify
the identity:

1.) tan (2 tan^-1 x) = 2 tan (tan^-1 x+tan^-1 x^3)

Many thanks!

 

[soroban]

Hello, Brandon!

Here's some help . . .

\(\displaystyle 1)\;\cos(2x)\,-\,\sin^2\left(\frac{x}{2}\right)\,+\,\frac{3}{4}\:=\:0\)

Use the identities: \(\displaystyle \:\begin{array}{ccc}\cos2\theta & \:=\: & 2\cos^2\theta\,-\,1 \\ \sin^2\left(\frac{\theta}{2}\right) & = & \frac{1\,-\,\cos\theta}{2} \end{array}\)

We have: \(\displaystyle \:\left[2\cos^2x\,-\,1\right] \,-\,\left[\frac{1\,-\,\cos x}{2}\right]\,+\,\frac{3}{4}\:=\:0\)

Multiply by 4: \(\displaystyle \:8cos^2x\,-\,4\,-\,2\,+\,2\cos x \,+\,3\:=\:0\)

We have a quadratic: \(\displaystyle \:8\cos^2x\,+\,2\cos x\,-\,3\:=\:0\)

. . which factors: \(\displaystyle \2\cos x\,-\,1)(4\cos x\,+\,3)\:=\:0\)


And we have two equations to solve:

. . \(\displaystyle 2\cos x\,-\,1\:=\:0\;\;\Rightarrow\;\;\cos x \:=\:\frac{1}{2}\;\;\Rightarrow\;\;\fbox{x\:=\:60^o,\:300^o}\)

. . \(\displaystyle 4\cos x\,+\,3\:=\:0\;\;\Rightarrow\;\;\cos x \:=\:-\frac{3}{4}\;\;\Rightarrow\;\;\fbox{x \:\approx\:138^o35',\;221^o25'}\)

\(\displaystyle 2)\;\cos(4x) \:= \sin(2x)\)

Use the identity: \(\displaystyle \:\cos(2\theta) \:=\:1\,-\,2\sin^2\theta\)

The equation becomes: \(\displaystyle \:1\,-\,2\sin^2(2x)\:=\:\sin(2x)\)

We have a quadratic: \(\displaystyle \:2\sin^2(2x)\,+\,\sin(2x)\,-\,1\:=\:0\)

. . which factors: \(\displaystyle \:[2\sin(2x)\,-\,1]\,[\sin(2x)\,+\,1] \:=\:0\)


And we have two equations to solve:

. . \(\displaystyle 2\sin(2x)\,-\,1\:=\:0\;\;\Rightarrow\;\;\sin(2x)\,=\,\frac{1}{2}\)
. . . . \(\displaystyle 2x\:=\:30^o,\:150^o,\:390^o,\:510^o\;\;\Rightarrow\;\;\fbox{x \:=\:15^o, \:75^o,\:195^o,\:255^o}\)

. . \(\displaystyle \sin(2x)\,+\,1\:=\:0\;\;\Rightarrow\;\;\sin(2x)\,=\,-1\)
. . . . \(\displaystyle 2x\,=\,270^o,\:630^o\;\;\Rightarrow\;\;\fbox{x\:=\:135^o,\:315^o}\)