<p>If sin theta= 4/5 and pi/2<theta<pi then tan theta=?
I got 4/3
but the answer is -4/3
Please show steps, why am I wrong????
Thanks, you're a lifesaver, may angels always bless your heart!</p>
<p>Well, first off, your parameters are π/2 and π, which means 90˚ and 180˚ if you prefer degrees. If you think of a unit circle, 90 degrees is on top of the circle, and a quarter way around is 90˚, so your theta is in quadrant two. Tangent can be thought of as the slope because it is sin/cos, and sintheta refers to the y value while cosinetheta refers to the x value, so if your theta is in the top left hand corner, it has to go down to the lower right hand corner, which means the slope is negative. Whatever steps you did gave you two values (I think), but you have to choose the negative one because tangent is negative in quadrant 2 (Also quadrant 4). I’ll post the steps in the next comment if no one else gets to it while I am doing it, but thinking of a unit circle really helps.</p>
<p>What you are forgetting is ASTC. Google the acronym on how to draw it if my explanation is confusing: draw your two axes, which give you your 4 quadrants. ASTC goes in the direction that quadrants are numbered, one letter in each quadrant.
These letters stand for which trig values are positive in which quardants: A for all, S for sin, T for tan, and C for cos. Since theta is between pi/2 and pi, you are looking at quadrant two. The letter in that quadrant is S, meaning that ONLY sin is positive in that quadrant. Hence, tan theta = -4/3
The acronym to remember this is All Students Take Calc</p>
<p>I’ll do two methods, unit circle way which is best for tests like these, and trigonometric equations and identities which I think is more thorough but not that great for standardized tests.</p>
<p>Your parameters are 90˚ and 180˚, so right away you should be thinking in the upper left hand corner. Sin is equal to opposite over hypotenuse. Your opposite side is the y value, because if you draw a triangle with its base at the π line, the opposite side is opposite the corner that touches the center of the circle. Then, after drawing that triangle, label your hypotenuse 5. In order to find the adjacent side, we use the pythagorean theorem, or a^2+b^2=c^2. Right away you should see this as a 3, 4, 5 right triangle, because 3^2+4^2=5^2. So, your x value is 3, but because it’s in the upper left hand corner, it’s technically negative 3. Tangent is opposite over adjacent, so you do 4/-3 which is -4/3.</p>
<p>While this took a while to type, you should practice this so that you can solve these types of problems very quickly. Also, try to start picturing it instead of drawing it. That would make it go even faster.</p>
<p>Now for trig identities and equations, which is just for fun and thoroughness but shouldn’t really be used on the ACT, lol.</p>
<p>sinø=4/5
sin^2ø=16/25
A trigonometric identity is 1-sin^2ø=cos^2ø
1-16/25 is 9/25.
cos^2ø=9/25
Taking the square root of that, cosø=+or-3/5
tanø=sinø/cosø
tanø= (4/5)/(3/5) or (4/5)/(-3/5)
You can divide by fractions by multiplying by the reciprocals, so your answers for tanø are (20/15) and (-20/15).
These simplify to (4/3) and (-4/3).
You just have to know that in quadrant 2, tangent is negative, so -4/3 is the right answer.</p>
<p>@Cosmological @mitec12 thanks a lot!</p>
<p>Trig hates you</p>
<p>jk I don’t know what Trig likes</p>