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If Theta = 13pi /6 What Does Sin Theta =

If sin θ = (-3/5) and π < θ < 3π/2, then what is tan θ?

Hi, I can see that people have come up with many different methods like using trigonometric identities like [math]sin^2 ({\theta})+ cos^2 ({\theta})= 1[/math] and then finding out the value of [math]tan {\theta}[/math]I will be explaining this question in a method which I think is the easiest and makes sense. I am going to be using 2 concepts, namely the pythagoras’ theorem and the signs of the 4 quadrants in the cartesian plane.Firstly, according to your question, [math]sin {\theta}= \dfrac{-3}{5}[/math] and [math]{\pi}< {\theta} < \dfrac{3{\pi}} {2}[/math]Now, we can consider a right angled triangle say [math]{\Delta}ABC [/math]where angle[math] ABC= {\theta} [/math]We can ignore the negative sign of [math]sin {\theta}[/math] and find out the value of your desired trigonometric function which is [math]tan {\theta}[/math]By applying pythagoras’ theorem, the unknown side is 4 units which is the side adjacent to angle [math]{\theta}[/math]Now we know that [math]tan {\theta}= \dfrac {opposite side}{adjacent side} [/math]We get [math]tan {\theta}= \dfrac{\pm {3}}{4}[/math]Here, the value of [math]tan {\theta} [/math]can be [math]{\pm}[/math] depending upon which quadrant it is present in.Now we can consider the second part of the question, where [math]{\pi}<{\theta} < \dfrac{3{\pi}}{2}[/math]This just implies that theta lies in the 3rd quadrant, henceIn the third quadrant, tan and cot are postive, you can remember these signs by knowing the acronym All students take chocolates where the initial letters give the positive sign of all functions, sin and its reciprocal, you get the idea…..Hence, your value of [math]tan {\theta}= \dfrac{+3}{4}[/math]If you just use the pythagoras theorem and the signs, you can practically get the answer in 2 lines.

Sin(2theta)=(1/2), 0<=theta<2pi?

Sin(2 theta) = (1/2), 0 <= theta < 2pi

As TC points out, this means 2 theta could equal pi/6 or 5 pi / 6. However, since theta can range up to 2 pi, 2 theta can range up to 4 pi, thus we must include the solutions 13 pi/6 and 17 pi / 6 as well.

Thus theta is any of the following:

pi/12
5pi/12
13pi/12
17pi/12

If sin(3 theta)=-1 and 0

sin(3x) = -1
then 3x can be
3x = 3pi/2,5pi/2, 7pi/2, 9pi/2 , 11pi/2, 13pi/2 and so on,
there fore
x = 3pi/6, 5pi/6, 7pi/6, 9pi/6, 11pi/6, 13pi/6 and so on
but solution
only 3pi/6, 5pi/6, 7pi/6, 9pi/6, 11pi/6 will be in the given range.

What can be the possible values for [math]\theta+\phi[/math] if [math]\sec2\theta = \tan\phi + \cot\phi[/math] ?

I'm supposing theeta = a & phi = bGiven: sec 2a = tan b + cot bTo find: possible values of ( a+ b)Since, sec 2a = tan b + cot b=> sec 2a = (tan² b +1) / tan b = sec²b / tan b=> sec 2a = 1/(sin b * cos b)=> 1/ cos 2a = 1/ (sin b * cos b) ………. (1)Now since, 2sin b *cos b = sin 2b=> sin b* cos b = sin 2b / 2By putting up this value in ….. (1)we get, 1/ cos 2a = 2/ sin 2b=> sin 2b / cos 2a = 2 /1=> the ratio of sin value to cos value = 2:1=> sin value is double & cos value is halfBut highest sin value is 1 …….. (2)●=> the highest cos value = 1/2 …….. (2)’ ●Lowest sin value = -1 ……… (3)●=> Lowest cos value = -1/2 …….. (3)’●Since, sin 2b = 1 , cos 2a = 1/2=> sin 2*45° = 1 , cos 2*30° = 1/2=> b= 45°, a = 30° ……….. (4)Again, since sin 90° = 1, & cos 300° = 1/2=> b= 45° , a= 150° ………….. (5)Similarly, since sin 270° = -1 & cos 120° = -1/2=> b= 135°, a= 60° ………….. (6)By (4), (5), & (6) we can say that( a + b) = 75° & 195°=> ( a+ b) lies between 75°( 5pi/12) to 195° ( 13pi/12)75° <, = (a+b) <, = 195°Or, 5pi/12 <,= (a+b) <,= 13pi/12

Solve equation sin 2 theta = square root of 3/2?

Let @ = theta.

sin 2@ = sqrt(3) / 2

The sine function equals sqrt(3) / 2 at the following angles measured in degrees:
....,-300, 60, 420, 780,1140,....,60 + 360k and
....,-240, 120, 480, 840, 1200,....,120 + 360k where k is an integer.

2@ = 60 + 360K and
2@ = 120 + 360K
divide both sides of each equation by 2

ANSWER:
in degrees
@ = 30 + 180K and
@ = 60 + 180K where K is an integer.

in radians
@ = pi / 6 + (pi)K and
@ = pi / 3 + (pi)K where K is an integer.

How can i find the sin and cos of 315 degrees?

315 degrees is 45 degrees in the 4th Q.
sin 45 = sqrt 2/2
cos 45 = sqrt 2/2
sin is - in Q4 so - sqrt 2/2
cos is + so sqrt 2/2

(13 pi/6)(180/pi) = 13(30) = 390
390 degrees is 30 degrees in Q1
sin 30 = 1/2
cos 30 = sqrt 3/2

How do I evaluate sin (23pi/12)?

Sin (23pi / 12) = sin (2pi - pi/12).Sin (pi/12) = sin (pi/4 - pi/6).Sin (a - b) = sinAcosB - sinBcosA.So,Sin (23pi/12) = sin (2pi) cos (pi/12) - sin (pi/12) cos (2pi).Notice that Sin (2pi) = 0 and cos (2pi) = 1.Sin (23pi/12) = - sin (pi/12).Sin (pi/12) = sin (pi/4 - pi/6) = sin(pi/4)cos(pi/6) - sin(pi/6)cos(pi/4).

How do you solve 1/2=sin(∅)?

sin(theta) = 1/2sin(alpha) = 1/2 (where alpha is an angle in the first quadrant)alpha = 30 degreesNow we have to consider the other 4 quadrants. We know Sin is positive so using our ASTC diagram we know theta is in quadrants 1 or 2.Therefore theta = 30 degrees, 180–30 degressSo 30, 150

How do I solve sin π/14 × sin 3π /14 × sin 5π/14?

Hope it's clear to you

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