heureka

i need the answer and solution preferably please.

if  \(\large{ \tan(120^\circ-x)=\dfrac{\sin(120^\circ) -\sin(x)} {\cos(120^\circ) -\cos(x)} }\), where \(\large{0^\circ < x < 180^\circ}\), compute \(x\)

\(\text{Let $ \mathbf{s} = \sin(120^\circ)= \sin(60^\circ)=\dfrac{\sqrt{3}} {2} $} \\ \text{Let $ \mathbf{c} = \cos(120^\circ)=-\cos(60^\circ)=-\dfrac{1} {2} $} \\ \)

LHS:

\(\begin{array}{|rcll|} \hline \tan(120^\circ-x) &=& \dfrac{\sin(120^\circ-x)} {\cos(120^\circ-x)} \\\\ &=& \dfrac{\sin(120^\circ)\cos(x)-\cos(120^\circ)\sin(x)} {\cos(120^\circ)\cos(x)+\sin(120^\circ)\sin(x)} \\\\ &=& \dfrac{s\cdot \cos(x)-c\cdot\sin(x)} {c\cdot\cos(x)+s\cdot\sin(x)} \\ \hline \end{array}\)

RHS:

\(\begin{array}{|rcll|} \hline \dfrac{\sin(120^\circ) -\sin(x)} {\cos(120^\circ) -\cos(x)} \\\\ &=& \dfrac{ 2\cos\left(\dfrac{120^\circ+x}{2}\right) \sin\left(\dfrac{120^\circ-x}{2}\right) } {-2\sin\left(\dfrac{120^\circ+x}{2}\right) \sin\left(\dfrac{120^\circ-x}{2}\right)} \\\\ &=& -\dfrac{ \cos\left(\dfrac{120^\circ+x}{2}\right) } { \sin\left(\dfrac{120^\circ+x}{2}\right) } \\\\ &=& -\dfrac{ \cos\left(60^\circ+\dfrac{x}{2}\right) } { \sin\left(60^\circ+\dfrac{x}{2}\right) } \\\\ &=& -\dfrac{\cos(60^\circ)\cos\left(\dfrac{x}{2}\right) -\sin(60^\circ)\sin\left(\dfrac{x}{2}\right) } {\sin(60^\circ)\cos\left(\dfrac{x}{2}\right) +\cos(60^\circ)\sin\left(\dfrac{x}{2}\right) } \\\\ &=& \dfrac{\sin(60^\circ)\sin\left(\dfrac{x}{2}\right) -\cos(60^\circ)\cos\left(\dfrac{x}{2}\right) } {\sin(60^\circ)\cos\left(\dfrac{x}{2}\right) +\cos(60^\circ)\sin\left(\dfrac{x}{2}\right) } \\\\ &=& \dfrac{s\cdot \sin\left(\dfrac{x}{2}\right) +c\cdot \cos\left(\dfrac{x}{2}\right) } {s\cdot \cos\left(\dfrac{x}{2}\right) -c\cdot \sin\left(\dfrac{x}{2}\right) } \\ \hline \end{array}\)

\(\begin{array}{|lcll|} \hline \dfrac{s\cdot \cos(x)-c\cdot\sin(x)} {c\cdot\cos(x)+s\cdot\sin(x)} = \dfrac{s\cdot \sin\left(\dfrac{x}{2}\right) + c\cdot \cos\left(\dfrac{x}{2}\right) } {s\cdot \cos\left(\dfrac{x}{2}\right) - c\cdot \sin\left(\dfrac{x}{2}\right) } \\\\ (s\cdot \cos(x)-c\cdot\sin(x)) \left(s\cdot \cos\left(\dfrac{x}{2}\right) - c\cdot \sin\left(\dfrac{x}{2}\right) \right) \\ = (c\cdot\cos(x)+s\cdot\sin(x)) \left( s\cdot \sin\left(\dfrac{x}{2}\right) + c\cdot \cos\left(\dfrac{x}{2}\right) \right) \\\\ s^2\cos(x)\cos\left(\dfrac{x}{2}\right) - sc\cos(x)\sin\left(\dfrac{x}{2}\right)- sc\sin(x)\cos\left(\dfrac{x}{2}\right)+c^2\sin(x)\sin\left(\dfrac{x}{2}\right) \\ =c^2\cos(x)\cos\left(\dfrac{x}{2}\right) + sc\cos(x)\sin\left(\dfrac{x}{2}\right)+ sc\sin(x)\cos\left(\dfrac{x}{2}\right)+s^2\sin(x)\sin\left(\dfrac{x}{2}\right) \\\\ \cos(x)\cos\left(\dfrac{x}{2}\right)(s^2-c^2)-\sin(x)\sin\left(\dfrac{x}{2}\right)(s^2-c^2)-2sc\cdot \cos(x)\sin\left(\dfrac{x}{2}\right)-2sc\cdot \sin(x)\cos\left(\dfrac{x}{2}\right) = 0 \\\\ (s^2-c^2)\left(\underbrace{ \cos(x)\cos\left(\dfrac{x}{2}\right)-\sin(x)\sin\left(\dfrac{x}{2}\right) }_{=\cos\left(x+\dfrac{x}{2}\right)} \right) -2sc\cdot\left(\underbrace{ \cos(x)\sin\left(\dfrac{x}{2}\right)+\sin(x)\cos\left(\dfrac{x}{2}\right) }_{=\sin\left(x+\dfrac{x}{2}\right)} \right) = 0 \\\\ (s^2-c^2)\cos\left(x+\dfrac{x}{2}\right) - 2sc\cdot \sin\left(x+\dfrac{x}{2}\right) = 0 \\\\ 2sc\cdot \sin\left(x+\dfrac{x}{2}\right) = (s^2-c^2)\cos\left(x+\dfrac{x}{2}\right) \\\\ 2sc\cdot \sin\left(\dfrac{3x}{2}\right) = (s^2-c^2)\cos\left(\dfrac{3x}{2}\right) \\\\ \tan\left(\dfrac{3x}{2}\right) = \dfrac{s^2-c^2} {2sc} \quad | \quad s^2-c^2 = \dfrac34-\dfrac14 = \dfrac12 \\\\ \tan\left(\dfrac{3x}{2}\right) = \dfrac{\dfrac12} {2sc} \quad | \quad 2sc = 2\dfrac{\sqrt{3}}{2}\left(-\dfrac12\right)=-\dfrac{\sqrt{3}}{2} \\\\ \tan\left(\dfrac{3x}{2}\right) = \dfrac{\dfrac12} {-\dfrac{\sqrt{3}}{2}} \\\\ \tan\left(\dfrac{3x}{2}\right) = -\dfrac{\sqrt{3}}{3} \\\\ \dfrac{3x}{2} = \arctan(-\dfrac{\sqrt{3}}{3}) +n\cdot 180^\circ \qquad n\in \mathbb{Z} \\\\ x= \dfrac{2}{3}\cdot \arctan\left(-\dfrac{\sqrt{3}}{3}\right) +\dfrac{2}{3}\cdot n\cdot 180^\circ \\\\ x= \dfrac{2}{3}\cdot (-30^\circ) + n\cdot 120^\circ \\\\ x= -20^\circ + n\cdot 120^\circ \qquad n = 1 \quad (0^\circ < x < 180^\circ) \\\\ x= -20^\circ + 120^\circ \\\\ \mathbf{x= 100^\circ } \\ \hline \end{array}\)

Find all real values of a for which the equation \(\mathbf{(x^2 + a)^2 + a = x}\) has four real roots.

My attempt:

I assume there are 4 distinct real roots. So there are 3 local maxima/minima.

We find the local maxima/minima by differentiation. Maxima/minima occur when f'(x) = 0

1. Differentiation:

\(\begin{array}{|rcll|} \hline y &=& (x^2 + a)^2 -x+a \\ y' &=& 2(x^2+a)\cdot 2x-1 \\ 0 &=& 4x(x^2+a) - 1 \\ 4x(x^2+a) &=& 1 \quad | \quad : 4 \\ x(x^2+a) &=& \frac14 \\ \mathbf{x^3+ax-\frac14} & \mathbf{=} & \mathbf{0} \\ \hline \end{array}\)

The zero set of discriminant of the cubic  \(\mathbf{Ax^{3}+Bx^{2}+Cx+D\,}\) has discriminant
  \(\mathbf{ B^{2}C^{2}-4AC^{3}-4B^{3}D-27A^{2}D^{2}+18ABCD\,.}\)
The discriminant is zero if and only if at least two roots are equal.
If the coefficients are real numbers, and the discriminant is not zero,
the discriminant is positive if the roots are three distinct real numbers,
and negative if there is one real root and two complex conjugate roots.

\(\begin{array}{|rcll|} \hline \mathbf{Ax^{3}+Bx^{2}+Cx+D\,} \\ \mathbf{x^3+ax-\frac14} & \mathbf{=} & \mathbf{0} \quad | \quad A=1,\ B=0,\ C=a,\ D=-\frac14 \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{ B^{2}C^{2}-4AC^{3}-4B^{3}D-27A^{2}D^{2}+18ABCD } &>& 0 \quad | \quad \text{there are three distinct real numbers} \\ -4a^3 -27\left(-\frac14\right)^2 &>& 0 \\ -4a^3 -\frac{27}{16} &>& 0 \\ -4a^3 &>& \frac{27}{16} \quad | \quad : -4 \\ a^3 &<& -\frac{27}{64} \\ a^3 &<& -\frac{3^3}{4^3} \\ \mathbf{ a } &\mathbf{<}& \mathbf{-\frac{3}{4}} \\ \hline \end{array}\)

\(\mathbf{(x^2 + a)^2 + a = x}\) has four distinct real roots if  \(\mathbf{ a<-\frac{3}{4}}\)

If a, b, c and d are positive real numbers such that
log_a  b= 8/9,
log_b  c= -(3/4),
log_c d = 2,
find the value of log_d (abc)

\(\begin{array}{|rclcl|} \hline \log_a(b) &=& \dfrac{\log_d(b)}{\log_d(a)} &=& \dfrac{\log_b(b)}{\log_b(a)} \\\\ & & \dfrac{\log_d(b)}{\log_d(a)} &=& \dfrac{\log_b(b)}{\log_b(a)} \quad | \quad \log_b(b) = 1 \\\\ & & \dfrac{\log_d(b)}{\log_d(a)} &=& \dfrac{1}{\log_b(a)} \quad | \quad \log_a(b)\log_b(a)=1 \\\\ & & \dfrac{\log_d(b)}{\log_d(a)} &=& \log_a(b) \\\\ & & \dfrac{\log_d(a)} {\log_d(b)}&=& \dfrac{1}{\log_a(b)} \\\\ & & \mathbf{\log_d(a)} & \mathbf{=} & \mathbf{ \dfrac{\log_d(b)}{\log_a(b)} } \qquad (1) \\ \hline \log_b(c) &=& \dfrac{\log_d(c)}{\log_d(b)} &=& \dfrac{\log_c(c)}{\log_c(b)} \\\\ & & \dfrac{\log_d(c)}{\log_d(b)} &=& \dfrac{\log_c(c)}{\log_c(b)} \quad | \quad \log_c(c) = 1 \\\\ & & \dfrac{\log_d(c)}{\log_d(b)} &=& \dfrac{1}{\log_c(b)} \quad | \quad \log_b(c)\log_c(b)=1 \\\\ & & \dfrac{\log_d(c)}{\log_d(b)} &=& \log_b(c) \\\\ & & \dfrac{\log_d(b)}{\log_d(c)} &=& \dfrac{1}{\log_b(c)} \\\\ & & \mathbf{\log_d(b)} &\mathbf{=}& \mathbf{\dfrac{\log_d(c)}{\log_b(c)}} \qquad (2) \\ \hline && \log_c(d) &=& \dfrac{\log_d(d)}{\log_d(c)} \quad | \quad \log_d(d) = 1 \\\\ && \log_c(d) &=& \dfrac{1}{\log_d(c)} \\\\ & & \mathbf{\log_d(c)} &\mathbf{=}& \mathbf{\dfrac{1}{\log_c(d)}} \qquad (3) \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \log_d(abc) &=& \log_d(a)+\log_d(b)+\log_d(c) \quad | \quad \mathbf{\log_d(a)=\dfrac{\log_d(b)}{\log_a(b)} } \qquad (1) \\ &=& \dfrac{\log_d(b)}{\log_a(b)}+\log_d(b)+\log_d(c) \\ &=& \log_d(b) \left( \dfrac{1}{\log_a(b)}+1 \right) +\log_d(c) \quad | \quad \mathbf{\log_d(b)=\dfrac{\log_d(c)}{\log_b(c)}} \qquad (2) \\ &=& \dfrac{\log_d(c)}{\log_b(c)} \left( \dfrac{1}{\log_a(b)}+1 \right) +\log_d(c) \\ &=&\log_d(c)\left( \dfrac{1}{\log_b(c)} \left( \dfrac{1}{\log_a(b)}+1 \right) + 1 \right) \quad | \quad \mathbf{\log_d(c)=\dfrac{1}{\log_c(d)}} \qquad (3) \\ \mathbf{\log_d(abc)} & \mathbf{=} & \mathbf{\dfrac{1}{\log_c(d)}\left( \dfrac{1}{\log_b(c)} \left( \dfrac{1}{\log_a(b)}+1 \right) + 1 \right)} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{\log_d(abc)} & \mathbf{=} & \mathbf{\dfrac{1}{\log_c(d)}\left( \dfrac{1}{\log_b(c)} \left( \dfrac{1}{\log_a(b)}+1 \right) + 1 \right)} \\\\ & = & \dfrac{1}{2}\left( \dfrac{1}{ -\dfrac{3}{4} } \left( \dfrac{1}{\dfrac{8}{9}}+1 \right) + 1 \right) \\\\ & = & \dfrac{1}{2}\left( \dfrac{-4}{3} \Big( \dfrac{1}{8}+1 \Big) + 1 \right) \\\\ & = & \dfrac{1}{2}\left( \dfrac{-4}{3} \left( \dfrac{17}{8} \right) + 1 \right) \\\\ & = & \dfrac{1}{2}\left( \dfrac{-17}{6} + 1 \right) \\\\ & = & \dfrac{1}{2}\left( \dfrac{-11}{6} \right) \\\\ & \mathbf{=} & -\mathbf{\dfrac{11}{12}} \\ \hline \end{array}\)

3)

3X^2 - X - 2

Answer i wrote: (3x - 2) (x + 1) no!

3X² - X - 2 = (3x + 2) (x - 1)

 For how many integers n between 1 and 11 (inclusive) n/12 is a repeating decimal?

\(\begin{array}{|r|r|r|r|r|} \hline & & & & & \text{prime number only } \\ & & & \text{factorize} & \text{prime numbers} & 2 \text{ or } 5 \\ n & \dfrac{n}{12} & \text{cancel} & \text{the denominator} & \text{in the denominator}& \text{will terminate} \\ \hline 1 & \dfrac{1}{12} & \dfrac{1}{12} & \dfrac{1}{2^2\cdot 3} & 2,\ 3 \\ \hline 2 & \dfrac{2}{12} & \dfrac{1}{6} & \dfrac{1}{2 \cdot 3} & 2,\ 3 \\ \hline 3 & \dfrac{3}{12} & \dfrac{1}{4} & \dfrac{1}{2^2} & 2 & \checkmark \Rightarrow \text{terminate} \\ \hline 4 & \dfrac{4}{12} & \dfrac{1}{3} & \dfrac{1}{3} & 3 \\ \hline 5 & \dfrac{5}{12} & \dfrac{5}{12} & \dfrac{1}{2^2\cdot 3} & 2,\ 3 \\ \hline 6 & \dfrac{6}{12} & \dfrac{1}{2} & \dfrac{1}{2} & 2 & \checkmark \Rightarrow \text{terminate} \\ \hline 7 & \dfrac{7}{12} & \dfrac{7}{12} & \dfrac{7}{2^2\cdot 3} & 2,\ 3 \\ \hline 8 & \dfrac{8}{12} & \dfrac{2}{3} & \dfrac{2}{3} & 3 \\ \hline 9 & \dfrac{9}{12} & \dfrac{3}{4} & \dfrac{3}{2^2} & 2 & \checkmark \Rightarrow \text{terminate} \\ \hline 10 & \dfrac{10}{12} & \dfrac{5}{6} & \dfrac{5}{2\cdot 3} & 2,\ 3 \\ \hline 11 & \dfrac{11}{12} & \dfrac{11}{12} & \dfrac{11}{2^2\cdot 3} & 2,\ 3 \\ \hline \end{array} \)

Hi Guest,

\(\begin{array}{|rcll|} \hline && \mathbf{\cos(195^\circ)} \\ &=& \cos(180^\circ +15^\circ)\\ &=& \underbrace{\cos(180^\circ)}_{=-1}\cos(15^\circ) - \underbrace{\sin(180^\circ)}_{=0}\sin(15^\circ)\\ &=& -\cos(15^\circ) - 0\cdot \sin(15^\circ)\\ &\mathbf{=}& \mathbf{-\cos(15^\circ)} \\\\ &=& -\cos(45^\circ - 30^\circ) \\ &=& -\Big( \underbrace{\cos(45^\circ)}_{=\dfrac{\sqrt{2}}{2}} \underbrace{\cos(30^\circ) }_{=\dfrac{\sqrt{3}}{2}} + \underbrace{\sin(45^\circ)}_{=\dfrac{\sqrt{2}}{2}} \underbrace{\sin(30^\circ)}_{=\dfrac{1}{2}}\Big) \\ &=& -\dfrac14 *\sqrt{2}*(\sqrt{3}+1) \\ \hline \end{array}\)

Thank you CPhill,

Sorry GA,

I corrected my typo:

\(\boxed{\dfrac{1}{k(k+1)} = \dfrac{1}{k} - \dfrac{1}{k+1} } \)

Determining the exact value of each trig expression.

Your calculation is correct.

\(\cos(195^\circ)=-\cos(15°) = -\dfrac14 *\sqrt{2}*(\sqrt{3}+1) \\ \tan(165^\circ)=-\tan(15°) = \sqrt{3} - 2\)

Reduction

\(\begin{array}{|rcll|} \hline x\cdot (100\ \%-23\ \%) &=& £57.50 \\ x\cdot(77\ \%) &=& £57.50 \\ x&=&\dfrac{ £57.50 }{77\ \%} \\ x&=&\dfrac{ £57.50 }{\dfrac{77}{100}} \\ x&=&\dfrac{ £57.50\cdot 100 }{77} \\ x&=&\dfrac{ £5750}{77} \\ x&=& £74.6753246753 \\ \hline \end{array}\)

The normal price is \(\approx £74.68\)