Co-functions

Hay,

I started with co-functions a little while back, I got the concept and understand the A + B = 90⁰,

But i got this one that threw me a curve-ball,

sin (90⁰ + x) = 2 cos²x

I spoke to my lecturer about this one and he said just to treat the second part as a normal square,

Where is what i tried...(Thinking)

sin (90⁰ + x) = 2[sin(90⁰-x)²]

sin (90⁰ + x) = 2[(90⁰-x)(90⁰-x)]

90⁰-x = 2[810⁰ - 180⁰x + x²]

90⁰-x = 1620⁰ - 360⁰x + 2x

2x² - 361⁰x + 1530⁰ = 0

Using quadratic formula:

.'. x = 3 + "root 12231" /4 or .'. x = 3 -"root 12231" /4

.'. x = 28.398 or .'. x = -27.145

Did i miss something here or look at it from a wrong angle?

sin(90+x)=cosx
therefore substitute sin(90+x)=cosx
and then solve the trigonometric equation cosx = 2 cos^2 (x) which is easy...

Thanks Minoanman,

I didn't know you where allowed to substitute,

I thought the point of the exercise was to determine x with only using co-functions,

sin (90⁰ + x) = 2 cos²x

cosx = 2cos²x

divide both side with cosx

.'. 1 = 2 cosx

.'. cosx = 1/2

.'. x = 60⁰; 90⁰; 270⁰; 300⁰

Quote:

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Originally Posted by MiloMan

Thanks Minoanman,

I didn't know you where allowed to substitute,

I thought the point of the exercise was to determine x with only using co-functions,

sin (90⁰ + x) = 2 cos²x

cosx = 2cos²x

divide both side with cosx

.'. 1 = 2 cosx

.'. cosx = 1/2

.'. x = 60⁰; 90⁰; 270⁰; 300⁰

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No, you can't divide both sides by $\displaystyle \displaystyle \begin{align*} \cos{(x)} \end{align*}$ because it's possible for (and in this case, some solutions ARE where) $\displaystyle \displaystyle \begin{align*} \cos{(x)} = 0 \end{align*}$. Instead you need to factorise...

$\displaystyle \displaystyle \begin{align*} \cos{(x)} &= 2\cos^2{(x)} \\ 0 &= 2\cos^2{(x)} - \cos{(x)} \\ 0 &= \cos{(x)} \left[ 2\cos{(x)} - 1 \right] \\ \cos{(x)} &= 0 \textrm{ or } 2\cos{(x)} - 1 = 0 \end{align*}$

Go from here.

Quote:

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Originally Posted by MiloMan

Thanks Minoanman,

I didn't know you where allowed to substitute,

I thought the point of the exercise was to determine x with only using co-functions,

sin (90⁰ + x) = 2 cos²x

cosx = 2cos²x

divide both side with cosx

.'. 1 = 2 cosx

.'. cosx = 1/2

.'. x = 60⁰; 90⁰; 270⁰; 300⁰

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How you could obtain 90 and 270 degrees from your solution? This is a flag that you did something incorrectly.

I must say i'm not sure where they got the 90 and 270 degree from,

I just copied it from the text book,

Doesn't 90 and 270 degree come from the fact the cos 90 and cos 270 = 0 and -0,

So if i put in 0 on both sides of the equation ( cosx = 2cos²x )

Then i sould be ok???

Quote:

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Originally Posted by MiloMan

I must say i'm not sure where they got the 90 and 270 degree from,

I just copied it from the text book,

Doesn't 90 and 270 degree come from the fact the cos 90 and cos 270 = 0 and -0,

So if i put in 0 on both sides of the equation ( cosx = 2cos²x )

Then i sould be ok???

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90 and 270 degrees are solutions to the problem. You could not get those angles from deviding both sides of your equation by cos(x). Follow the method provided by Prove It.

Quote:

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Originally Posted by MiloMan

sin (90⁰ + x) = 2[(90⁰-x)(90⁰-x)]

90⁰-x = 2[810⁰ - 180⁰x + x²]

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It is the sine function that is squared, not the argument.
$\displaystyle sin^2(x - x_0) \neq sin( (x - x_0 )^2 )$
and then you just dropped the sine function. You did this again a couple of steps later.

-Dan

Since you say you understand how "co-functions" are related you could do it this way: sin(90+ x)= sin(90- (-x))= cos(-x). And since cosine is an even function, cos(-x)= cos(x), so
sin(90+ x)= 2cos^2(x) is the same as cos(x)= 2 cos^2(x). cos(x)= 0 is one solution which gives $\displaystyle x= \pi/2+ n\pi$ for any integer n. If cos(x) is not 0, we can divide both sides by cos(x) giving 1= 2cos(x) or cos(x)= 1/2. From that, [tex]x= \pi/3+ 2n\pi[tex] and $\displaystyle x= 2n\pi- \pi/3$.