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# Tutor profile: Dalian S.

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Dalian S.
Tutor for 4 years
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## Questions

### Subject:Trigonometry

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Question:

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

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Dalian S.

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

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Question:

What's 20\% of 500?

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Dalian S.

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

### Subject:Algebra

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Question:

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

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Dalian S.

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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