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Popular Trigonometry >

sin(θ)+cos(θ)= 7/5 ,sin(2θ)

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Solution

sin(θ)+cos(θ)=57​,sin(2θ)

Solution

NoSolutionforθ∈R
Solution steps
sin(θ)+cos(θ)=57​,sin(2θ)
Rewrite using trig identities
sin(θ)+cos(θ)
sin(θ)+cos(θ)=2​sin(θ+4π​)
sin(θ)+cos(θ)
Rewrite as=2​(2​1​sin(θ)+2​1​cos(θ))
Use the following trivial identity: cos(4π​)=2​1​Use the following trivial identity: sin(4π​)=2​1​=2​(cos(4π​)sin(θ)+sin(4π​)cos(θ))
Use the Angle Sum identity: sin(s+t)=sin(s)cos(t)+cos(s)sin(t)=2​sin(θ+4π​)
=2​sin(θ+4π​)
2​sin(θ+4π​)=57​
Divide both sides by 2​
2​sin(θ+4π​)=57​
Divide both sides by 2​2​2​sin(θ+4π​)​=2​57​​
Simplify
2​2​sin(θ+4π​)​=2​57​​
Simplify 2​2​sin(θ+4π​)​:sin(θ+4π​)
2​2​sin(θ+4π​)​
Cancel the common factor: 2​=sin(θ+4π​)
Simplify 2​57​​:1072​​
2​57​​
Apply the fraction rule: acb​​=c⋅ab​=52​7​
Rationalize 52​7​:1072​​
52​7​
Multiply by the conjugate 2​2​​=52​2​72​​
52​2​=10
52​2​
Apply radical rule: a​a​=a2​2​=2=5⋅2
Multiply the numbers: 5⋅2=10=10
=1072​​
=1072​​
sin(θ+4π​)=1072​​
sin(θ+4π​)=1072​​
sin(θ+4π​)=1072​​
Apply trig inverse properties
sin(θ+4π​)=1072​​
General solutions for sin(θ+4π​)=1072​​sin(x)=a⇒x=arcsin(a)+2πn,x=π−arcsin(a)+2πnθ+4π​=arcsin(1072​​)+2πn,θ+4π​=π−arcsin(1072​​)+2πn
θ+4π​=arcsin(1072​​)+2πn,θ+4π​=π−arcsin(1072​​)+2πn
Solve θ+4π​=arcsin(1072​​)+2πn:θ=arcsin(52​7​)+2πn−4π​
θ+4π​=arcsin(1072​​)+2πn
Simplify arcsin(1072​​)+2πn:arcsin(52​7​)+2πn
arcsin(1072​​)+2πn
1072​​=52​7​
1072​​
Factor 10:2⋅5
Factor 10=2⋅5
=2⋅572​​
Cancel 2⋅572​​:2​⋅57​
2⋅572​​
Apply radical rule: na​=an1​2​=221​=2⋅57⋅221​​
Apply exponent rule: xbxa​=xb−a1​21221​​=21−21​1​=5⋅2−21​+17​
Subtract the numbers: 1−21​=21​=5⋅221​7​
Apply radical rule: an1​=na​221​=2​=52​7​
=2​⋅57​
=arcsin(52​7​)+2πn
θ+4π​=arcsin(52​7​)+2πn
Move 4π​to the right side
θ+4π​=arcsin(52​7​)+2πn
Subtract 4π​ from both sidesθ+4π​−4π​=arcsin(52​7​)+2πn−4π​
Simplify
θ+4π​−4π​=arcsin(52​7​)+2πn−4π​
Simplify θ+4π​−4π​:θ
θ+4π​−4π​
Add similar elements: 4π​−4π​=0
=θ
Simplify arcsin(52​7​)+2πn−4π​:arcsin(52​7​)+2πn−4π​
arcsin(52​7​)+2πn−4π​
=arcsin(1072​​)+2πn−4π​
1072​​=52​7​
1072​​
Factor 10:2⋅5
Factor 10=2⋅5
=2⋅572​​
Cancel 2⋅572​​:2​⋅57​
2⋅572​​
Apply radical rule: na​=an1​2​=221​=2⋅57⋅221​​
Apply exponent rule: xbxa​=xb−a1​21221​​=21−21​1​=5⋅2−21​+17​
Subtract the numbers: 1−21​=21​=5⋅221​7​
Apply radical rule: an1​=na​221​=2​=52​7​
=2​⋅57​
=arcsin(52​7​)+2πn−4π​
Could not simplify further=arcsin(52​7​)+2πn−4π​
θ=arcsin(52​7​)+2πn−4π​
θ=arcsin(52​7​)+2πn−4π​
θ=arcsin(52​7​)+2πn−4π​
Solve θ+4π​=π−arcsin(1072​​)+2πn:θ=π−arcsin(52​7​)+2πn−4π​
θ+4π​=π−arcsin(1072​​)+2πn
Simplify π−arcsin(1072​​)+2πn:π−arcsin(52​7​)+2πn
π−arcsin(1072​​)+2πn
1072​​=52​7​
1072​​
Factor 10:2⋅5
Factor 10=2⋅5
=2⋅572​​
Cancel 2⋅572​​:2​⋅57​
2⋅572​​
Apply radical rule: na​=an1​2​=221​=2⋅57⋅221​​
Apply exponent rule: xbxa​=xb−a1​21221​​=21−21​1​=5⋅2−21​+17​
Subtract the numbers: 1−21​=21​=5⋅221​7​
Apply radical rule: an1​=na​221​=2​=52​7​
=2​⋅57​
=π−arcsin(52​7​)+2πn
θ+4π​=π−arcsin(52​7​)+2πn
Move 4π​to the right side
θ+4π​=π−arcsin(52​7​)+2πn
Subtract 4π​ from both sidesθ+4π​−4π​=π−arcsin(52​7​)+2πn−4π​
Simplify
θ+4π​−4π​=π−arcsin(52​7​)+2πn−4π​
Simplify θ+4π​−4π​:θ
θ+4π​−4π​
Add similar elements: 4π​−4π​=0
=θ
Simplify π−arcsin(52​7​)+2πn−4π​:π−arcsin(52​7​)+2πn−4π​
π−arcsin(52​7​)+2πn−4π​
=π−arcsin(1072​​)+2πn−4π​
1072​​=52​7​
1072​​
Factor 10:2⋅5
Factor 10=2⋅5
=2⋅572​​
Cancel 2⋅572​​:2​⋅57​
2⋅572​​
Apply radical rule: na​=an1​2​=221​=2⋅57⋅221​​
Apply exponent rule: xbxa​=xb−a1​21221​​=21−21​1​=5⋅2−21​+17​
Subtract the numbers: 1−21​=21​=5⋅221​7​
Apply radical rule: an1​=na​221​=2​=52​7​
=2​⋅57​
=π−arcsin(52​7​)+2πn−4π​
Could not simplify further=π−arcsin(52​7​)+2πn−4π​
θ=π−arcsin(52​7​)+2πn−4π​
θ=π−arcsin(52​7​)+2πn−4π​
θ=π−arcsin(52​7​)+2πn−4π​
θ=arcsin(52​7​)+2πn−4π​,θ=π−arcsin(52​7​)+2πn−4π​
Solutions for the range sin(2θ)NoSolutionforθ∈R

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tan(x)= 5/9tan(t)=-(sqrt(11))/5 ,cos(t)>0,sin(t)tan(θ+20)tan(90-3θ)=12csc(x)+7/(cos(x))=0csc(θ)= 13/6

Frequently Asked Questions (FAQ)

  • What is the general solution for sin(θ)+cos(θ)= 7/5 ,sin(2θ) ?

    The general solution for sin(θ)+cos(θ)= 7/5 ,sin(2θ) is No Solution for θ\in\mathbb{R}
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