anonymous
  • anonymous
If Sin θ = 3/5, find the value of other all trig functions.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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Nnesha
  • Nnesha
what's the definition of sin ?
anonymous
  • anonymous
Umm idk
Nnesha
  • Nnesha
alright thats fine what do you know about trig ratios ? if cos = adjacent over hypotenuse then sin = ?

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anonymous
  • anonymous
Opp/hyp
Nnesha
  • Nnesha
correct its based on right triangle |dw:1449203112435:dw| opposite and the adjacent sides depends on the angle adjacent side is the side that touches the angle ( next to the angle )
Nnesha
  • Nnesha
for example if the angle is y|dw:1449203229016:dw| adjacent side of angle y would be GL
Nnesha
  • Nnesha
draw a right triangle and use the given information sin theta = 3/5 |dw:1449203330468:dw| where would you put 3 and 5 ? which side is 3 and which one is 5 ?
anonymous
  • anonymous
Would the answer be cos=+-4/5
anonymous
  • anonymous
Can you help me with a couple more questions?
anonymous
  • anonymous
1. Simplify : sin B + (cos^2 B/ Sin B) 2. Find the value of sin θ + cos θ cot θ - csc θ 3. Simplify: sec A – ( tan2 A/ Sec A)
Nnesha
  • Nnesha
you need to write `all` trig ratios that means 6 trigg ratios
anonymous
  • anonymous
So how do I write all the trig ratios
anonymous
  • anonymous
Sin=opposite/hypotenuse Sin=3/5 From Pythagorean theorem, 5^2-3^2-adjacent side^2 25-9=16 Adjacent side=4 cos=adj./hyp., =4/5 Tan=opp./ adj., =3/4 sec=1/cos,=5/4 cosec=1/sin, =5/3 cotan=1/tan, =4/3
Nnesha
  • Nnesha
good job. next time just post the link instead of copy pasting full paragraph.
anonymous
  • anonymous
How do I do these ? 18. Simplify: sin B + (cos2B/ Sin B) 19. Find the value of sin θ + cos θ cot θ - csc θ. 20. Simplify: sec A – ( tan2 A/ Sec A) ​​​
Nnesha
  • Nnesha
is it cos^2 B or cos (2B)
anonymous
  • anonymous
Cos^2
anonymous
  • anonymous
Simplify: sin B + (cos^2B/ Sin B)
Nnesha
  • Nnesha
alright \[\large\rm \sin B + \frac{\cos^2B}{sinB}\] find the common denomiantor
Nnesha
  • Nnesha
denominator
anonymous
  • anonymous
Sine B is the common denominator
Nnesha
  • Nnesha
right so how would you write it as single fraction
anonymous
  • anonymous
2sinB
Nnesha
  • Nnesha
just 2 sinb remember you need to get the common denominator for sin B to do that you should multiply top and sin B by the common denominator example \[\large\rm a+\frac{c}{d} \rightarrow \frac{a*d+c}{d}\]
anonymous
  • anonymous
1-2sin^2B
Nnesha
  • Nnesha
how would write \[\rm \sin B +\frac{\cos^2B}{sinB} \]as a single fraction (common denominator ) look at the example
anonymous
  • anonymous
1 + sin b - cos b = 1 + 2 sin b/2 cos b/2 - (1 - 2 sin² b/2) = 2 sin b/2 cos b/2 + 2 sin² b/2 = 2 sin b/2 (cos b/2 + sin b/2 ) So then 1 +sin b -cos b =0 2 sin b/2 (cos b/2 + sin b/2 )=0
Nnesha
  • Nnesha
i have no idea what u did there the comman denominator is sin b so to write it as single fraction you should multiply \[\rm \sin B*\frac{\sin B}{\sin B}+\frac{\cos^2B}{\sin B} \]simplify
Nnesha
  • Nnesha
now the denominators are the same u can write it as single fraction
Nnesha
  • Nnesha
and then use the identity sin^2x+cos^2x=1

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