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solve the trigonometric function
tan∅ × sin∅ × cos∅ + cos2∅

Sagot :

Answer:

cos²Θ

Step-by-step explanation:

simplify the expression using the identities

tanΘ = [tex]\frac{sin0}{cos0}[/tex]

cos2Θ = 1 - 2sin²Θ

cos²Θ = 1 - sin²Θ

then

tanΘ × sinΘ × cosΘ + cos2Θ

= [tex]\frac{sin0}{cos0}[/tex] × sinΘ × cosΘ + 1 - 2sin²Θ ( cancel cosΘ on numerator/ denominator )

= sinΘ × sinΘ + 1 - 2sin²Θ

= sin²Θ + 1 - 2sin²Θ

= 1 - sin²Θ

= cos²Θ

semsee45

Answer:

[tex]\cos^2(\theta)[/tex]

Step-by-step explanation:

Trig identities used:

[tex]\cos(2\theta)=\cos^2(\theta)-\sin^2(\theta)[/tex]

[tex]\tan(\theta)=\dfrac{\sin(\theta)}{\cos(\theta)}[/tex]

Therefore,

[tex]\tan(\theta)\times\sin(\theta)\times\cos(\theta)+\cos(2\theta)[/tex]

[tex]=\dfrac{\sin(\theta)}{\cos(\theta)}\times\sin(\theta)\times\cos(\theta)+\cos(2\theta)[/tex]

[tex]=\sin^2(\theta) +\cos(2\theta)[/tex]

[tex]=\sin^2(\theta) +\cos^2(\theta)-\sin^2(\theta)[/tex]

[tex]=\cos^2(\theta)[/tex]