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Tutor profile: Dalian S.

Inactive
Dalian S.
Tutor for 4 years
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Questions

Subject: Trigonometry

TutorMe
Question:

If $\sin\theta=\frac{3}{5}$, and $\theta$ is an acute angle, then find $\sin 3\theta$.

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Dalian S.
Answer:

Firstly since $\theta$ is acute, then $cos\theta$ must be positive. To calculate $\cos\theta$ we use the rule,\\ $\cos^2\theta+\sin^2\theta=1$\\ Substituting $\sin\theta=\frac{1}{3}$ and solving for $\cos\theta\ge 0$, we find,\\ $cos\theta=\frac{4}{5}$.\\ Now we can use the sine compound angle formula to expand $\sin 3\theta$,\\ $\sin 3\theta=\sin\left(2\theta +\theta\right)=\sin 2\theta\cos\theta+\cos 2\theta\sin\theta$\\ which can further be simplified to, $\sin 3\theta = 2\sin\theta\cos\theta\cos\theta+\left(2\cos^2\theta-1\right)\sin\theta$, Finally we substitute $\sin\theta=\frac{3}{5}$ and $\cos\theta=\frac{4}{5}$ to obtain $\sin 3\theta$.

Subject: Basic Math

TutorMe
Question:

What's 20\% of 500?

Inactive
Dalian S.
Answer:

$20\% of 500 = 500/100 \times 20 = 5 \times 20 = 100$

Subject: Algebra

TutorMe
Question:

Solve for $x$, $3x^4-5x^3+2x^2+3x-5=-2$

Inactive
Dalian S.
Answer:

Take a common factor of $x^2$ from the first 2 terms,\\ $x^2(3x-5)$+2x^2+(3x-5)=-2$,\\ Now take a common factor of $3x-5$,\\ $(3x-5)(x^2+1)+2x^2=-2$,\\ Add 2 to both sides of the equation,\\ $(3x-5)(x^2+1)+2x^2+2=-2+2$,\\ Take out common factor again,\\ $(3x-5)(x^2+1)+2(x^2+1)=0$,\\ Again we take out common factor,\\ $(x^2+1)(3x-5+2)=0$\\ From this we obtain solutions, $3x-5+2=0$, which results in $3x=3$ and simplifies to $x=1$.\\ The other equations results in $x^2=-1$ or equivalently, $x=--i$ or $x=i$.

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