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ratelena [41]
3 years ago
8

1

Mathematics
1 answer:
devlian [24]3 years ago
5 0

Answer:

-1

Step-by-step explanation:

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If cos theta = 1/4 and 0 degrees, find tan theta
Zina [86]
What? Sorry can’t understand
8 0
3 years ago
Help me with trigonometry
poizon [28]

Answer:

See below

Step-by-step explanation:

It has something to do with the<em> </em><u><em>Weierstrass substitution</em></u>, where we have

$\int\, f(\sin(x), \cos(x))dx = \int\, \dfrac{2}{1+t^2}f\left(\dfrac{2t}{1+t^2}, \dfrac{1-t^2}{1+t^2} \right)dt$

First, consider the double angle formula for tangent:

\tan(2x)= \dfrac{2\tan(x)}{1-\tan^2(x)}

Therefore,

\tan\left(2 \cdot\dfrac{x}{2}\right)= \dfrac{2\tan(x/2)}{1-\tan^2(x/2)} = \tan(x)=\dfrac{2t}{1-t^2}

Once the double angle identity for sine is

\sin(2x)= \dfrac{2\tan(x)}{1+\tan^2(x)}

we know \sin(x)=\dfrac{2t}{1+t^2}, but sure,  we can derive this formula considering the double angle identity

\sin(x)= 2\sin\left(\dfrac{x}{2}\right)\cos\left(\dfrac{x}{2}\right)

Recall

\sin \arctan t = \dfrac{t}{\sqrt{1 + t^2}} \text{ and } \cos \arctan t = \dfrac{1}{\sqrt{1 + t^2}}

Thus,

\sin(x)= 2 \left(\dfrac{t}{\sqrt{1 + t^2}}\right) \left(\dfrac{1}{\sqrt{1 + t^2}}\right) = \dfrac{2t}{1 + t^2}

Similarly for cosine, consider the double angle identity

Thus,

\cos(x)=  \left(\dfrac{1}{\sqrt{1 + t^2}}\right)^2- \left(\dfrac{t}{\sqrt{1 + t^2}}\right)^2 = \dfrac{1}{t^2+1}-\dfrac{t^2}{t^2+1} =\dfrac{1-t^2}{1+t^2}

Hence, we showed \sin(x) \text { and } \cos(x)

======================================================

5\cos(x) =12\sin(x) +3, x \in [0, 2\pi ]

Solving

5\,\overbrace{\frac{1-t^2}{1+t^2}}^{\cos(x)} = 12\,\overbrace{\frac{2t}{1+t^2}}^{\sin(x)}+3

\implies \dfrac{5-5t^2}{1+t^2}= \dfrac{24t}{1+t^2}+3 \implies  \dfrac{5-5t^2 -24t}{1+t^2}= 3

\implies 5-5t^2-24t=3\left(1+t^2\right) \implies -8t^2-24t+2=0

t = \dfrac{-(-24)\pm \sqrt{(-24)^2-4(-8)\cdot 2}}{2(-8)} = \dfrac{24\pm 8\sqrt{10}}{-16} =  \dfrac{3\pm \sqrt{10}}{-2}

t=-\dfrac{3+\sqrt{10}}{2}\\t=\dfrac{\sqrt{10}-3}{2}

Just note that

\tan\left(\dfrac{x}{2}\right) =  \dfrac{3\pm 8\sqrt{10}}{-2}

and  \tan\left(\dfrac{x}{2}\right) is not defined for x=k\pi , k\in\mathbb{Z}

6 0
3 years ago
PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE HELP ME!!!!!!!! I HAVE 2 QUESTIONS THATS ALL. I WILL GIVE POINTS, FAN, MEDAL. PLEASE I
Arte-miy333 [17]
I think it would be a also but im not good at math
7 0
4 years ago
What is tan(B)?<br> B<br> 5<br> 13<br> 5<br> 12<br> ?<br> tan(B) =<br> =<br> Enter
SOVA2 [1]

Answer:

tan B = 12/5

Step-by-step explanation:

Since this is a right  triangle, we can use trig functions

tan theta = opp side/ adj side

tan B = 12/5

3 0
2 years ago
PLEASE HURRY, thanks!
forsale [732]
The answer to the problem is C 2^12 the reason is because 16^3 is 4,096 and
2^7=128
2^11=2,048
2^12=4,096 which is what we want so this is the answer
2^64=18,447,744,073,709,551,616 which is not even close to what we want at all

So the answer is C 2^12
8 0
4 years ago
Read 2 more answers
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