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Scorpion4ik [409]
1 year ago
14

Find the six trig function values of the angle 240*Show all work, do not use calculator

Mathematics
1 answer:
-BARSIC- [3]1 year ago
5 0

Solution:

Given:

240^0

To get sin 240 degrees:

240 degrees falls in the third quadrant.

In the third quadrant, only tangent is positive. Hence, sin 240 will be negative.

sin240^0=sin(180+60)

Using the trigonometric identity;

sin(x+y)=sinx\text{ }cosy+cosx\text{ }siny

Hence,

\begin{gathered} sin(180+60)=sin180cos60+cos180sin60 \\ sin180=0 \\ cos60=\frac{1}{2} \\ cos180=-1 \\ sin60=\frac{\sqrt{3}}{2} \\  \\ Thus, \\ sin180cos60+cos180sin60=0(\frac{1}{2})+(-1)(\frac{\sqrt{3}}{2}) \\ sin180cos60+cos180sin60=0-\frac{\sqrt{3}}{2} \\ sin180cos60+cos180sin60=-\frac{\sqrt{3}}{2} \\  \\ Hence, \\ sin240^0=-\frac{\sqrt{3}}{2} \end{gathered}

To get cos 240 degrees:

240 degrees falls in the third quadrant.

In the third quadrant, only tangent is positive. Hence, cos 240 will be negative.

cos240^0=cos(180+60)

Using the trigonometric identity;

cos(x+y)=cosx\text{ }cosy-sinx\text{ }siny

Hence,

\begin{gathered} cos(180+60)=cos180cos60-sin180sin60 \\ sin180=0 \\ cos60=\frac{1}{2} \\ cos180=-1 \\ sin60=\frac{\sqrt{3}}{2} \\  \\ Thus, \\ cos180cos60-sin180sin60=-1(\frac{1}{2})-0(\frac{\sqrt{3}}{2}) \\ cos180cos60-sin180sin60=-\frac{1}{2}-0 \\ cos180cos60-sin180sin60=-\frac{1}{2} \\  \\ Hence, \\ cos240^0=-\frac{1}{2} \end{gathered}

To get tan 240 degrees:

240 degrees falls in the third quadrant.

In the third quadrant, only tangent is positive. Hence, tan 240 will be positive.

tan240^0=tan(180+60)

Using the trigonometric identity;

tan(180+x)=tan\text{ }x

Hence,

\begin{gathered} tan(180+60)=tan60 \\ tan60=\sqrt{3} \\  \\ Hence, \\ tan240^0=\sqrt{3} \end{gathered}

To get cosec 240 degrees:

\begin{gathered} cosec\text{ }x=\frac{1}{sinx} \\ csc240=\frac{1}{sin240} \\ sin240=-\frac{\sqrt{3}}{2} \\  \\ Hence, \\ csc240=\frac{1}{\frac{-\sqrt{3}}{2}} \\ csc240=-\frac{2}{\sqrt{3}} \\  \\ Rationalizing\text{ the denominator;} \\ csc240=-\frac{2}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}} \\  \\ Thus, \\ csc240^0=-\frac{2\sqrt{3}}{3} \end{gathered}

To get sec 240 degrees:

\begin{gathered} sec\text{ }x=\frac{1}{cosx} \\ sec240=\frac{1}{cos240} \\ cos240=-\frac{1}{2} \\  \\ Hence, \\ sec240=\frac{1}{\frac{-1}{2}} \\ sec240=-2 \\  \\ Thus, \\ sec240^0=-2 \end{gathered}

To get cot 240 degrees:

\begin{gathered} cot\text{ }x=\frac{1}{tan\text{ }x} \\ cot240=\frac{1}{tan240} \\ tan240=\sqrt{3} \\  \\ Hence, \\ cot240=\frac{1}{\sqrt{3}} \\  \\ Rationalizing\text{ the denominator;} \\ cot240=\frac{1}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}} \\  \\ Thus, \\ cot240^0=\frac{\sqrt{3}}{3} \end{gathered}

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If you're using the app, try seeing this answer through your browser:  brainly.com/question/3000160

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Find the derivative of

\mathsf{y=\ell n(sec\,x+tan\,x)}\\\\\\ \mathsf{y=\ell n\!\left(\dfrac{1}{cos\,x}+\dfrac{sin\,x}{cos\,x} \right )}\\\\\\ \mathsf{y=\ell n\!\left(\dfrac{1+sin\,x}{cos\,x} \right )}


You can treat  y  as a composite function of  x:

\left\{\! \begin{array}{l} \mathsf{y=\ell n\,u}\\\\ \mathsf{u=\dfrac{1+sin\,x}{cos\,x}} \end{array} \right.


so use the chain rule to differentiate  y:

\mathsf{\dfrac{dy}{dx}=\dfrac{dy}{du}\cdot \dfrac{du}{dx}}\\\\\\ \mathsf{\dfrac{dy}{dx}=\dfrac{d}{du}(\ell n\,u)\cdot \dfrac{d}{dx}\!\left(\dfrac{1+sin\,x}{cos\,x}\right)}


The first derivative is  1/u, and the second one can be evaluated by applying the quotient rule:

\mathsf{\dfrac{dy}{dx}=\dfrac{1}{u}\cdot \dfrac{\frac{d}{dx}(1+sin\,x)\cdot cos\,x-(1+sin\,x)\cdot \frac{d}{dx}(cos\,x)}{(cos\,x)^2}}\\\\\\ \mathsf{\dfrac{dy}{dx}=\dfrac{1}{u}\cdot \dfrac{(0+cos\,x)\cdot cos\,x-(1+sin\,x)\cdot (-\,sin\,x)}{(cos\,x)^2}}


Multiply out those terms in parentheses:

\mathsf{\dfrac{dy}{dx}=\dfrac{1}{u}\cdot \dfrac{cos\,x\cdot cos\,x+(sin\,x+sin\,x\cdot sin\,x)}{(cos\,x)^2}}\\\\\\ \mathsf{\dfrac{dy}{dx}=\dfrac{1}{u}\cdot \dfrac{cos^2\,x+sin\,x+sin^2\,x}{(cos\,x)^2}}\\\\\\ \mathsf{\dfrac{dy}{dx}=\dfrac{1}{u}\cdot \dfrac{(cos^2\,x+sin^2\,x)+sin\,x}{(cos\,x)^2}\qquad\quad (but~~cos^2\,x+sin^2\,x=1)}\\\\\\ \mathsf{\dfrac{dy}{dx}=\dfrac{1}{u}\cdot \dfrac{1+sin\,x}{(cos\,x)^2}}


Substitute back for  \mathsf{u=\dfrac{1+sin\,x}{cos\,x}:}

\mathsf{\dfrac{dy}{dx}=\dfrac{1}{~\frac{1+sin\,x}{cos\,x}~}\cdot \dfrac{1+sin\,x}{(cos\,x)^2}}\\\\\\ \mathsf{\dfrac{dy}{dx}=\dfrac{cos\,x}{1+sin\,x}\cdot \dfrac{1+sin\,x}{(cos\,x)^2}}


Simplifying that product, you get

\mathsf{\dfrac{dy}{dx}=\dfrac{1}{1+sin\,x}\cdot \dfrac{1+sin\,x}{cos\,x}}\\\\\\ \mathsf{\dfrac{dy}{dx}=\dfrac{1}{cos\,x}}


∴     \boxed{\begin{array}{c}\mathsf{\dfrac{dy}{dx}=sec\,x} \end{array}}\quad\longleftarrow\quad\textsf{this is the answer.}


I hope this helps. =)


Tags:  <em>derivative composite function logarithmic logarithm log trigonometric trig secant tangent sec tan chain rule quotient rule differential integral calculus</em>

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