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Hi, I'm currently taking pre-cal, but I just don't understand how to simplify, solve and varify using the trigonometric functions and the many identities that I find confusing. For example--> (verifying if it is an identity)-->I)Sin^3θ-cos^3= (1+sinθcosθ)(sinθ-cosθ)
and-->(solving the equation for 0°≤ X ≤180°)-->i)secθ = 1+tanθ
-->ii)4sin^2θ-4sinθ+1=0
My teacher doesn't really know how to explain to the students regarding this.
Thank you
Thank you
sin^3(θ)-cos^3(Θ)= (1+sin(θ)cos(θ))(sin(θ)-cos(θ))
In the difference of cubes, x^3-y^3, it is equal to (x-y)(x²+xy+y²)
Applying this, sin^3(Θ)-cos^3(Θ) =
(sin(Θ)-cos(Θ))(sin²(Θ)+sin(Θ)cos(Θ)+cos²(Θ)).
It is known that sin²(Θ)+ cos²(Θ)=1, so this become (sin(Θ)-cos(Θ))(1+ sin(Θ)cos(Θ)).
sec(θ) = 1+tan(θ),
To solve sec(θ) = 1+tan(θ), I would first convert to sin(θ) and cos(θ) by knowing that sec(θ)=1/cos(θ) and tan(θ)=sin(θ)/cos(θ). Put these expressions in and multiply the entire equation by cos(θ).
4sin^2θ-4sinθ+1=0
For the equation 4sin^2θ-4sinθ+1=0, remember that 4x^2-4x+1 factors into (2x-1)². Replace the x with sin(Θ) and you’ve got the answer. All that needs to be done is to solve 2sin(Θ)=0.
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