Sin 270

If F is the composition of two differentiable functions f\left(u\right) and u=g\left(x\right), that is, if F\left(x\right)=f\left(g\left(x\right)\right), then the derivative of F is the derivative of f with respect to u times the derivative of g with respect to x, that is, \frac(g)\left(x\right).

Trigonometric ratios of 270 degree minus theta is one of the branches of ASTC formula in trigonometry. Trigonometric-ratios of 270 degree minus theta are given below. sin (270 ° - θ) = - cos θ cos (270 ° -θ) = - sin θ tan (270 ° -θ) = cot θ csc (270 ° -θ) = - sec θ sec (270 ° -θ) = - cos θ cot (270 ° -θ) = tan θ Let us see, how the trigonometric ratios of 270 degree minus theta are determined. To know that, first we have to understand ASTC formula. The ASTC formula can be remembered easily using the following phrases. "All Sliver Tea Cups" or "All Students Take Calculus" ASTC formula has been explained clearly in the figure given below. From the above picture, it is very clear that the angle 270 degree minus theta falls in the third quadrant. In the third quadrant (270 ° degree minus theta), tan and cot are positive and other trigonometric ratios are negative. Important Conversions When we have the angles 90 ° and 270 ° in the trigonometric ratios in the form of (9 0 ° + θ) (9 0 ° - θ) (27 0 ° + θ) (270 ° -θ) We have to do the following conversions, sin θ cos θ tan θ cot θ csc θ sec θ For example, sin (270 ° + θ) = - cos θ cos (90 ° - θ) = sin θ For the angles 0 ° or 360 ° and 180 °, we should not make the above conversions. Evaluation of Trigonometric Ratios 270 Degree Minus Theta Problem 1 : Evaluate : sin (27 0 ° - θ) Solution : To evaluate sin (27 0 ° - θ), we have to consider the following important points. (i) (27 0 ° - θ) will fall in the III rd quadrant. (ii) Wh...

I'm not new to trigonometry, but this question always bothers me. As it is in $$ \sin(\theta)=\frac $$ We know that the sum of the angles in a triangle is 180 degrees, which means each angle is less than 180 degrees, so how, for example, $\sin 270$ is possible? Edit After Joe posted MathWorld has already explained both definitions. I kept the question because of the brilliant answers below. $\begingroup$ This is a classic case of a mathematical idea being defined one way, and then extended beyond the original definition in a mathematically consistent and mathematically fruitful way. It can take a bit of getting used to, but as you progress in mathematics you will find it happens quite often. It is testimony to the fact that mathematical ideas, if they are good ones, can make connections between situations which appear to be very different. $\endgroup$ You are correct. If by $\sin \theta$ we mean $\frac$. So once again, sine becomes more abstract, but more powerful. Although the word ‘trigonometry’ originally meant ‹triangle measurement›, trigonometry is really about much more than just triangles. The formula $ \operatorname $ is correct when talking about angles in a right triangle, but that is only a small part of the ways in which angles appear in geometry, and geometry itself is only part of what the sine is good for. So it's a place to start, but there is much more to the sine than that. The two comments that have appeared under your question so far (by Deepak and run...

We know that, sin (90° + θ) = cos θ cos (90° + θ) = - sin θ tan (90° + θ) = - cot θ csc (90° + θ) = sec θ sec ( 90° + θ) = - csc θ cot ( 90° + θ) = - tan θ and sin (180° + θ) = - sin θ cos (180° + θ) = - cos θ tan (180° + θ) = tan θ csc (180° + θ) = -csc θ sec (180° + θ) = - sec θ cot (180° + θ) = cot θ Using the above proved results we will prove all six trigonometrical ratios of (180° - θ). sin (270° + θ) = sin [1800 + 90° + θ] = sin [1800 + (90° + θ)] = - sin (90° + θ), [since sin (180° + θ) = - sin θ] Therefore, sin (270° + θ) = - cos θ , [since sin (90° + θ) = cos θ] cos (270° + θ) = cos [1800 + 90° + θ] = cos [I 800 + (90° + θ)] = - cos (90° + θ), [since cos (180° + θ) = - cos θ] Therefore, cos (270° + θ) = sin θ , [since cos (90° + θ) = - sin θ] tan ( 270° + θ) = tan [1800 + 90° + θ] = tan [180° + (90° + θ)] = tan (90° + θ), [since tan (180° + θ) = tan θ] Therefore, tan (270° + θ) = - cot θ , [since tan (90° + θ) = - cot θ] csc (270° + θ) = \(\frac\) ● Trigonometric Functions • • • Reciprocal Relations of Trigonometric Ratios • • • • • • • • Trig Ratio Problems • Proving Trigonometric Ratios • • • • • • • • • • • • • • Trigonometrical Ratios of (- θ) • • • Trigonometrical Ratios of (180° + θ) • Trigonometrical Ratios of (180° - θ) • Trigonometrical Ratios of (270° + θ) • T • Trigonometrical Ratios of (360° + θ) • • • • • Trigonometric Functions of any Angles • • 11 and 12 Grade Math From Trigonometrical Ratios of (270° + θ) to HOME PAGE

I'm not new to trigonometry, but this question always bothers me. As it is in $$ \sin(\theta)=\frac $$ We know that the sum of the angles in a triangle is 180 degrees, which means each angle is less than 180 degrees, so how, for example, $\sin 270$ is possible? Edit After Joe posted MathWorld has already explained both definitions. I kept the question because of the brilliant answers below. $\begingroup$ This is a classic case of a mathematical idea being defined one way, and then extended beyond the original definition in a mathematically consistent and mathematically fruitful way. It can take a bit of getting used to, but as you progress in mathematics you will find it happens quite often. It is testimony to the fact that mathematical ideas, if they are good ones, can make connections between situations which appear to be very different. $\endgroup$ You are correct. If by $\sin \theta$ we mean $\frac$. So once again, sine becomes more abstract, but more powerful. Although the word ‘trigonometry’ originally meant ‹triangle measurement›, trigonometry is really about much more than just triangles. The formula $ \operatorname $ is correct when talking about angles in a right triangle, but that is only a small part of the ways in which angles appear in geometry, and geometry itself is only part of what the sine is good for. So it's a place to start, but there is much more to the sine than that. The two comments that have appeared under your question so far (by Deepak and run...

If F is the composition of two differentiable functions f\left(u\right) and u=g\left(x\right), that is, if F\left(x\right)=f\left(g\left(x\right)\right), then the derivative of F is the derivative of f with respect to u times the derivative of g with respect to x, that is, \frac(g)\left(x\right).

We know that, sin (90° + θ) = cos θ cos (90° + θ) = - sin θ tan (90° + θ) = - cot θ csc (90° + θ) = sec θ sec ( 90° + θ) = - csc θ cot ( 90° + θ) = - tan θ and sin (180° + θ) = - sin θ cos (180° + θ) = - cos θ tan (180° + θ) = tan θ csc (180° + θ) = -csc θ sec (180° + θ) = - sec θ cot (180° + θ) = cot θ Using the above proved results we will prove all six trigonometrical ratios of (180° - θ). sin (270° + θ) = sin [1800 + 90° + θ] = sin [1800 + (90° + θ)] = - sin (90° + θ), [since sin (180° + θ) = - sin θ] Therefore, sin (270° + θ) = - cos θ , [since sin (90° + θ) = cos θ] cos (270° + θ) = cos [1800 + 90° + θ] = cos [I 800 + (90° + θ)] = - cos (90° + θ), [since cos (180° + θ) = - cos θ] Therefore, cos (270° + θ) = sin θ , [since cos (90° + θ) = - sin θ] tan ( 270° + θ) = tan [1800 + 90° + θ] = tan [180° + (90° + θ)] = tan (90° + θ), [since tan (180° + θ) = tan θ] Therefore, tan (270° + θ) = - cot θ , [since tan (90° + θ) = - cot θ] csc (270° + θ) = \(\frac\) ● Trigonometric Functions • • • Reciprocal Relations of Trigonometric Ratios • • • • • • • • Trig Ratio Problems • Proving Trigonometric Ratios • • • • • • • • • • • • • • Trigonometrical Ratios of (- θ) • • • Trigonometrical Ratios of (180° + θ) • Trigonometrical Ratios of (180° - θ) • Trigonometrical Ratios of (270° + θ) • T • Trigonometrical Ratios of (360° + θ) • • • • • Trigonometric Functions of any Angles • • 11 and 12 Grade Math From Trigonometrical Ratios of (270° + θ) to HOME PAGE

Trigonometric ratios of 270 degree minus theta is one of the branches of ASTC formula in trigonometry. Trigonometric-ratios of 270 degree minus theta are given below. sin (270 ° - θ) = - cos θ cos (270 ° -θ) = - sin θ tan (270 ° -θ) = cot θ csc (270 ° -θ) = - sec θ sec (270 ° -θ) = - cos θ cot (270 ° -θ) = tan θ Let us see, how the trigonometric ratios of 270 degree minus theta are determined. To know that, first we have to understand ASTC formula. The ASTC formula can be remembered easily using the following phrases. "All Sliver Tea Cups" or "All Students Take Calculus" ASTC formula has been explained clearly in the figure given below. From the above picture, it is very clear that the angle 270 degree minus theta falls in the third quadrant. In the third quadrant (270 ° degree minus theta), tan and cot are positive and other trigonometric ratios are negative. Important Conversions When we have the angles 90 ° and 270 ° in the trigonometric ratios in the form of (9 0 ° + θ) (9 0 ° - θ) (27 0 ° + θ) (270 ° -θ) We have to do the following conversions, sin θ cos θ tan θ cot θ csc θ sec θ For example, sin (270 ° + θ) = - cos θ cos (90 ° - θ) = sin θ For the angles 0 ° or 360 ° and 180 °, we should not make the above conversions. Evaluation of Trigonometric Ratios 270 Degree Minus Theta Problem 1 : Evaluate : sin (27 0 ° - θ) Solution : To evaluate sin (27 0 ° - θ), we have to consider the following important points. (i) (27 0 ° - θ) will fall in the III rd quadrant. (ii) Wh...